Welcome to HSLC Guru! This page presents complete ASSEB Class 12 Physics Chapter 2 — Electrostatic Potential and Capacitance — Question Answer in English Medium. The notes are aligned with the Assam State School Education Board (ASSEB) Higher Secondary curriculum and cover every concept from electric potential and equipotential surfaces to capacitance, dielectrics, combinations of capacitors and the energy stored in a capacitor. Each exercise problem has a clean step-by-step solution, every formula appears in proper KaTeX, and three SVG diagrams illustrate the key apparatus. Read straight through, or jump to the section you need.
Chapter Summary
Chapter 2 builds the language of energy for electrostatics. While Chapter 1 described the electric field as a vector quantity acting on charges, here we introduce a scalar quantity — the electric potential — which often makes calculations far easier. Once potential is defined we naturally meet equipotential surfaces, the relation $\vec{E} = -\nabla V$, the potential energy of a system of charges, and the role of conductors and dielectrics. The chapter then turns to capacitors: devices that store electrical energy by holding equal and opposite charges on two conductors. We derive the capacitance of a parallel-plate capacitor, learn the rules for series and parallel combinations, study the effect of inserting a dielectric slab, and finally calculate the energy stored and the energy density of the electric field.
The electric potential at a point is the work done by an external agent in bringing a unit positive test charge from infinity to that point against the electrostatic force, without changing its kinetic energy:
$$V(\vec{r}) = \frac{W_{\infty\to r}}{q_0} = -\int_{\infty}^{\vec{r}} \vec{E}\cdot d\vec{l}.$$
Potential is a scalar, measured in volt (V), and the potential difference $V_A – V_B$ equals the work done per unit charge in moving a positive test charge from $B$ to $A$. For a single point charge $q$ in vacuum,
$$V(r) = \frac{1}{4\pi\epsilon_0}\,\frac{q}{r},$$
where the reference is taken at infinity. For a system of point charges potentials add algebraically because superposition holds. The potential energy of an assembled system of $N$ charges is
$$U = \frac{1}{4\pi\epsilon_0}\sum_{i An equipotential surface is one on which $V$ is constant; no work is done in moving a charge over it, so $\vec{E}$ is always perpendicular to such a surface and points from higher to lower potential. The pointwise link is $$\vec{E} = -\nabla V \quad\text{or, in 1D,}\quad E = -\frac{dV}{dr}.$$ Inside a conductor in electrostatic equilibrium $\vec{E}=0$ and $V$ is constant; charge resides only on the surface and the field just outside is $\sigma/\epsilon_0$ normal to the surface. Dielectrics are non-conducting materials that polarise in an external field; the dielectric constant $K$ (or relative permittivity $\epsilon_r$) measures how much the field is reduced inside the slab and how much the capacitance is enhanced. A capacitor is two conductors carrying charges $+Q$ and $-Q$ at potential difference $V$. Its capacitance is $C = Q/V$, measured in farad (F). For a parallel-plate capacitor with plate area $A$, separation $d$ and vacuum between the plates, $C = \epsilon_0 A/d$; with a dielectric of constant $K$ filling the gap, $C = K\epsilon_0 A/d$. In series the reciprocals add; in parallel the capacitances add. The energy stored equals the work done to charge it, $$U = \tfrac{1}{2}CV^{2} = \frac{Q^{2}}{2C} = \tfrac{1}{2}QV,$$ and is interpreted as energy stored in the electric field with density $u = \tfrac{1}{2}\epsilon_0 E^{2}$. Diagram 1 — Parallel-plate capacitor. Two flat conducting plates, area $A$, separation $d$, carry equal and opposite charges. The field between the plates is uniform; outside the plates it is essentially zero (ignoring fringe fields). Diagram 2 — Series and parallel combinations. In series the same charge $Q$ flows through every capacitor and the voltages add; in parallel the same voltage $V$ appears across every capacitor and the charges add. Diagram 3 — Equipotential surfaces. For a positive point charge equipotential surfaces are concentric spheres; field lines are radial and perpendicular to them. For a uniform field, equipotentials are parallel planes perpendicular to the field. Q1. Two charges $5\times10^{-8}\,\text{C}$ and $-3\times10^{-8}\,\text{C}$ are located $16\,\text{cm}$ apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Answer: Let $A$ carry $q_1=5\times10^{-8}\,\text{C}$ and $B$ carry $q_2=-3\times10^{-8}\,\text{C}$, separated by $d=0.16\,\text{m}$. Take a point $P$ at distance $x$ from $A$ on the line $AB$. Setting $V_A+V_B=0$: $$\frac{1}{4\pi\epsilon_0}\!\left[\frac{q_1}{x}+\frac{q_2}{d-x}\right]=0\;\Rightarrow\;\frac{5\times10^{-8}}{x}=\frac{3\times10^{-8}}{0.16-x}.$$ Solving, $5(0.16-x)=3x$, so $x=0.10\,\text{m}=10\,\text{cm}$ from the positive charge (between the two). For a point outside the line beyond $B$, take distance $y$ from $A$ with $y>d$: $$\frac{5\times10^{-8}}{y}+\frac{-3\times10^{-8}}{y-0.16}=0\;\Rightarrow\;5(y-0.16)=3y\;\Rightarrow\;y=0.40\,\text{m}.$$ Hence the potential is zero at $10\,\text{cm}$ and $40\,\text{cm}$ from the positive charge, both measured along the line $AB$ on the side of the negative charge. Q2. A regular hexagon of side $10\,\text{cm}$ has a charge $5\,\mu\text{C}$ at each of its vertices. Calculate the potential at the centre of the hexagon. Answer: The centre of a regular hexagon is at a distance equal to the side $r=0.10\,\text{m}$ from every vertex. Each charge contributes $V_1=kq/r$, so the six contributions add directly: $$V = 6\cdot\frac{1}{4\pi\epsilon_0}\frac{q}{r}=\frac{6\times(9\times10^{9})\times(5\times10^{-6})}{0.10}=2.7\times10^{6}\,\text{V}.$$ The potential at the centre is therefore $V\approx 2.7\times10^{6}\,\text{V}$. Q3. Two charges $2\,\mu\text{C}$ and $-2\,\mu\text{C}$ are placed at points $A$ and $B$, $6\,\text{cm}$ apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface? Answer: (a) The locus of points equidistant from $A$ and $B$ is the plane that perpendicularly bisects the line $AB$. On this plane $V_A+V_B=k(q/r)+k(-q/r)=0$, so it is the equipotential of zero potential. (b) On this plane the contributions to $\vec E$ along the line $AB$ from the two equal-and-opposite charges add (both pointing from $+$ to $-$), while components perpendicular to $AB$ cancel. Hence the field is everywhere directed from $A$ to $B$, i.e. normal to the equipotential plane. Q4. A spherical conductor of radius $12\,\text{cm}$ has a charge $1.6\times10^{-7}\,\text{C}$ distributed uniformly on its surface. What is the electric field (a) inside the sphere, (b) just outside the sphere, (c) at a point $18\,\text{cm}$ from the centre? Answer: (a) Inside a charged conductor in equilibrium $E=0$. (b) Just outside, the conductor behaves as a point charge at the centre: $$E=\frac{1}{4\pi\epsilon_0}\frac{q}{R^{2}}=\frac{(9\times10^{9})(1.6\times10^{-7})}{(0.12)^{2}}=1.0\times10^{5}\,\text{N C}^{-1}.$$ (c) At $r=0.18\,\text{m}$, $$E=\frac{(9\times10^{9})(1.6\times10^{-7})}{(0.18)^{2}}\approx4.4\times10^{4}\,\text{N C}^{-1},$$ directed radially outward. Q5. A parallel-plate capacitor with air between the plates has a capacitance of $8\,\text{pF}$. What will be the capacitance if the distance between the plates is reduced by half and the space between is filled with a substance of dielectric constant $6$? Answer: Initially $C_0=\epsilon_0 A/d=8\,\text{pF}$. After halving the gap and inserting the dielectric, $$C’=\frac{K\epsilon_0 A}{d/2}=2K\,C_0=2\times6\times 8=96\,\text{pF}.$$ The new capacitance is $\mathbf{96\,\text{pF}}$. Q6. Three capacitors each of capacitance $9\,\text{pF}$ are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a $120\,\text{V}$ supply? Answer: (a) For three equal capacitors in series, $1/C_s=3/C$, so $C_s=C/3=9/3=\mathbf{3\,\text{pF}}$. (b) In series the same charge $Q=C_sV=3\,\text{pF}\times120\,\text{V}=360\,\text{pC}$ flows through each, so each has potential difference $V_1=Q/C=360/9=\mathbf{40\,\text{V}}$, and indeed $3\times40=120\,\text{V}$ as required. Q7. Three capacitors of capacitances $2\,\text{pF}$, $3\,\text{pF}$ and $4\,\text{pF}$ are connected in parallel. (a) What is the total capacitance? (b) Determine the charge on each if the combination is connected to a $100\,\text{V}$ supply. Answer: (a) In parallel $C_p=C_1+C_2+C_3=2+3+4=\mathbf{9\,\text{pF}}$. (b) Each capacitor has the same voltage $V=100\,\text{V}$, so $Q=CV$ gives $Q_1=200\,\text{pC},\ Q_2=300\,\text{pC},\ Q_3=400\,\text{pC}$. Total charge $Q=900\,\text{pC}$, equal to $C_p V$. Q8. In a parallel-plate capacitor with air between the plates, each plate has area $6\times10^{-3}\,\text{m}^{2}$ and the plates are $3\,\text{mm}$ apart. Calculate the capacitance. If the capacitor is connected to a $100\,\text{V}$ supply, what is the charge on each plate? Answer: $C=\dfrac{\epsilon_0 A}{d}=\dfrac{(8.85\times10^{-12})(6\times10^{-3})}{3\times10^{-3}}=1.77\times10^{-11}\,\text{F}\approx 17.7\,\text{pF}$. Charge on each plate: $Q=CV=1.77\times10^{-11}\times 100=1.77\times10^{-9}\,\text{C}\approx1.8\,\text{nC}$. Q9. Explain what would happen if in the capacitor of Q8 a $3\,\text{mm}$ thick mica sheet (of dielectric constant $K=6$) were inserted between the plates, (a) while the voltage supply remained connected, (b) after the supply was disconnected. Answer: Inserting the slab changes $C$ to $C’=K C = 6\times 17.7\approx 106.2\,\text{pF}$. (a) Supply still connected. $V$ remains $100\,\text{V}$, so the new charge is $Q’=C’V = 106.2\times 10^{-12}\times 100 = 1.06\times10^{-8}\,\text{C}\approx 10.6\,\text{nC}$. The extra charge is supplied by the battery. (b) Supply disconnected. $Q$ is fixed at $1.77\,\text{nC}$. The voltage drops to $V’=Q/C’=1.77\times10^{-9}/(106.2\times10^{-12})\approx 16.7\,\text{V}$, i.e. it is reduced by the factor $K=6$. Q10. A $12\,\text{pF}$ capacitor is connected to a $50\,\text{V}$ battery. How much electrostatic energy is stored in the capacitor? Answer: $$U=\tfrac{1}{2}CV^{2}=\tfrac{1}{2}(12\times10^{-12})(50)^{2}=1.5\times10^{-8}\,\text{J}.$$ The capacitor stores $1.5\times10^{-8}\,\text{J}=15\,\text{nJ}$ of electrostatic energy. Q11. A $600\,\text{pF}$ capacitor is charged by a $200\,\text{V}$ supply. It is then disconnected from the supply and connected to another uncharged $600\,\text{pF}$ capacitor. How much electrostatic energy is lost in the process? Answer: Initial energy stored on the first capacitor: $$U_i=\tfrac{1}{2}CV^{2}=\tfrac{1}{2}(600\times10^{-12})(200)^{2}=1.2\times10^{-5}\,\text{J}.$$ When connected in parallel to an identical uncharged capacitor, charge redistributes equally; the common voltage becomes $V’=V/2=100\,\text{V}$ and total capacitance $1200\,\text{pF}$: $$U_f=\tfrac{1}{2}(1200\times10^{-12})(100)^{2}=6.0\times10^{-6}\,\text{J}.$$ Energy lost $\Delta U = U_i-U_f = 1.2\times10^{-5}-0.6\times10^{-5}=\mathbf{6.0\times10^{-6}\,\text{J}}$. Half the original energy is dissipated as heat and electromagnetic radiation in the connecting wires. Q12. Define the SI unit of electric potential. Answer: The volt (V). One volt is the potential difference between two points such that one joule of work is done in moving one coulomb of charge from one point to the other against the electrostatic force: $1\,\text{V}=1\,\text{J/C}$. Q13. Why is the work done in moving a charge over an equipotential surface zero? Answer: Because $W=q(V_A-V_B)$ and $V_A=V_B$ on an equipotential surface, giving $W=0$. Equivalently, the displacement is perpendicular to $\vec E$, so $\vec F\cdot d\vec l=0$. Q14. Show that the electric field is perpendicular to an equipotential surface. Answer: If $\vec E$ had a component along the surface, moving a unit positive charge in that direction would do non-zero work, contradicting $V=$ constant on the surface. Hence $\vec E$ must be perpendicular to every equipotential surface. Q15. Derive the expression $V=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r}$ for the potential due to a point charge. Answer: The field at distance $r’$ from $q$ is $E=kq/r’^{2}$ along $\hat r$. Bringing a unit positive test charge from infinity along the radial line: $$V(r)=-\int_{\infty}^{r}\!E\,dr’=-\int_{\infty}^{r}\!\frac{kq}{r’^{2}}\,dr’=\frac{kq}{r}=\frac{1}{4\pi\epsilon_0}\frac{q}{r}.$$ Q16. Obtain the potential at a point on the axis of an electric dipole. Answer: For a dipole of moment $\vec p$ with charges $\pm q$ separated by $2a$, on the axis at distance $r$ from the centre, $$V_{\text{axial}}=\frac{kq}{r-a}-\frac{kq}{r+a}=\frac{kq\,(2a)}{r^{2}-a^{2}}\xrightarrow{r\gg a}\frac{1}{4\pi\epsilon_0}\frac{p}{r^{2}}.$$ Q17. Show that the potential at a point on the equatorial line of a dipole is zero. Answer: Any point on the equatorial line is equidistant from the two equal and opposite charges, so $V=kq/r-kq/r=0$. Q18. Define capacitance and state its SI unit. Answer: Capacitance $C$ of a conductor (or capacitor) is the charge required to raise its potential by unit volt: $C=Q/V$. The SI unit is the farad (F), $1\,\text{F}=1\,\text{C/V}$. Q19. Derive the capacitance of a parallel-plate capacitor with air between the plates. Answer: For two large plates of area $A$ separated by $d$, the surface charge density is $\sigma=Q/A$. The uniform field between the plates is $E=\sigma/\epsilon_0$, and the potential difference is $V=Ed=\sigma d/\epsilon_0=Qd/(\epsilon_0 A)$. Hence $$C=\frac{Q}{V}=\frac{\epsilon_0 A}{d}.$$ Q20. Show that with a dielectric slab of constant $K$ filling the gap, $C’=KC$. Answer: The dielectric reduces the field by a factor $K$, so $E’=E/K$ and $V’=V/K$ for the same $Q$. Therefore $C’=Q/V’=K(Q/V)=KC=K\epsilon_0 A/d$. Q21. Derive an expression for the energy stored in a charged capacitor. Answer: Suppose at some stage the charge on the capacitor is $q$ and its voltage $q/C$. The work done to add $dq$ is $dW=(q/C)\,dq$. Integrating from $0$ to $Q$: $$U=\int_0^{Q}\frac{q}{C}\,dq=\frac{Q^{2}}{2C}=\tfrac{1}{2}CV^{2}=\tfrac{1}{2}QV.$$ Q22. Obtain the energy density in the field of a parallel-plate capacitor. Answer: Using $U=\tfrac{1}{2}CV^{2}=\tfrac{1}{2}(\epsilon_0 A/d)(Ed)^{2}=\tfrac{1}{2}\epsilon_0 E^{2}(Ad)$. The volume of the field region is $Ad$, so the energy per unit volume is $$u=\frac{U}{Ad}=\tfrac{1}{2}\epsilon_0 E^{2}.$$ Q23. A $4\,\mu\text{F}$ capacitor is charged by a $200\,\text{V}$ supply. How much energy is stored? Answer: $U=\tfrac12 CV^{2}=\tfrac12(4\times10^{-6})(200)^{2}=0.08\,\text{J}=80\,\text{mJ}$. Q24. Two capacitors of $3\,\mu\text{F}$ and $6\,\mu\text{F}$ are joined in series. Find the equivalent capacitance. Answer: $\dfrac{1}{C}=\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}$, so $C=2\,\mu\text{F}$. Q25. What happens to the capacitance of an isolated charged capacitor when (i) plate separation is doubled, (ii) a dielectric of constant $K$ is inserted? Answer: (i) $C\propto 1/d$, so capacitance becomes half. (ii) $C\to KC$, so capacitance becomes $K$ times the original. Q26. Why must conductors used as plates of a capacitor be very close together? Answer: Because $C=\epsilon_0 A/d$ is inversely proportional to the separation $d$. Smaller $d$ gives larger capacitance for a given plate area. Q27. Three identical capacitors of $C=4\,\mu\text{F}$ are connected as: two in parallel, the combination in series with the third. Find the total capacitance. Answer: Parallel of two: $4+4=8\,\mu\text{F}$. In series with $4\,\mu\text{F}$: $\dfrac{1}{C_t}=\dfrac{1}{8}+\dfrac{1}{4}=\dfrac{3}{8}$, so $C_t=\dfrac{8}{3}\,\mu\text{F}\approx 2.67\,\mu\text{F}$. Q28. The potential at the surface of a charged conducting sphere of radius $R$ carrying charge $Q$ is $V_0$. Find $V$ at distance $r>R$. Answer: Outside the sphere it behaves like a point charge: $V(r)=kQ/r=V_0(R/r)$. Q29. Write two differences between conductors and dielectrics. Answer: (i) Conductors have free electrons that move under any field; dielectrics have only bound charges that polarise. (ii) Inside a conductor in equilibrium $E=0$; inside a dielectric $E$ is reduced but not zero, $E_{\text{in}}=E_0/K$. Q30. If the electric potential in a region is $V(x)=5x^{2}+10x-9\,\text{V}$, find the electric field at $x=1\,\text{m}$. Answer: $E_x=-\dfrac{dV}{dx}=-(10x+10)$. At $x=1$, $E_x=-20\,\text{V m}^{-1}$, i.e. $20\,\text{V m}^{-1}$ in the $-x$ direction. Electric potential energy as a path-independent quantity. Because the electrostatic force is conservative, the work done in moving a charge between two points depends only on the endpoints, not on the path. Hence one can define a scalar $V$ at every point with the property $W_{A\to B}=q(V_A-V_B)$. The line integral $\oint\vec E\cdot d\vec l=0$ around any closed path is the differential statement of the same fact, and is one of Maxwell’s electrostatic equations. Once $V$ is known everywhere, $\vec E$ follows by differentiation. Why a capacitor stores charge. When a conductor receives charge it must raise its own potential — that is what limits how much charge it can hold at a given voltage. By placing a second, oppositely charged conductor very close, we induce charges that lower the original conductor’s potential, so much more charge can be stored at the same supply voltage. The ratio $C=Q/V$ measures this charge-storing ability. The unit, the farad, is enormous: $1\,\text{F}$ would require an isolated sphere of radius about $9\times10^{9}\,\text{m}$. Real capacitors are usually rated in $\mu\text{F}$, $\text{nF}$ or $\text{pF}$. Effect of a dielectric in detail. When a dielectric is placed in an external field $\vec E_0$, the molecules polarise and produce a polarisation density $\vec P$ that points from $-$ to $+$ within each molecule. At the dielectric’s two faces this polarisation produces bound surface charges $-\sigma_p$ and $+\sigma_p$ that partly cancel the free charges on the capacitor plates, reducing the net field to $\vec E=\vec E_0/K$. Since $V=Ed$, the potential difference also drops to $V_0/K$ for the same $Q$, so $C=Q/V=KC_0$. The dielectric thus enhances capacitance by a factor $K$, from $\sim 1.0006$ for air to $\sim 80$ for water and several thousand for ferroelectric ceramics. Capacitor with partial dielectric. If a slab of thickness $t$ ($t $$C=\frac{\epsilon_0 A}{d – t\!\left(1-\dfrac{1}{K}\right)}.$$ For $t\to d$ this reduces to $K\epsilon_0 A/d$, and for $K\to\infty$ (a metallic slab of thickness $t$) it reduces to $\epsilon_0 A/(d-t)$ — i.e. the metal slab effectively shortens the gap by its own thickness. Energy stored as field energy. The energy supplied by the battery while charging is not “stored on the plates”; it resides in the electric field between them. The energy-density formula $u=\tfrac12\epsilon_0 E^{2}$ is general — even a region of vacuum carrying a field stores energy at this density. This is the basis on which electromagnetic waves carry energy through empty space. Q31. Derive an expression for the electric potential at any point due to an electric dipole and discuss the special cases of axial and equatorial points. Answer: Consider an electric dipole consisting of two charges $-q$ and $+q$ separated by a distance $2a$, with dipole moment $\vec p = q(2\vec a)$ pointing from $-q$ to $+q$. Let $P$ be the field point at distance $r$ from the centre $O$ of the dipole, with the line $OP$ making an angle $\theta$ with the axis of the dipole. Let the distances of $P$ from $+q$ and $-q$ be $r_1$ and $r_2$ respectively. $$V_P = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_1}-\frac{q}{r_2}\right).$$ For $r\gg a$, geometry gives $r_1\approx r-a\cos\theta$ and $r_2\approx r+a\cos\theta$. Substituting and keeping leading order in $a/r$: $$V_P = \frac{q}{4\pi\epsilon_0}\frac{2a\cos\theta}{r^{2}}=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^{2}}=\frac{1}{4\pi\epsilon_0}\frac{\vec p\cdot\hat r}{r^{2}}.$$ Special cases. On the axial line ($\theta=0$ or $\pi$), $\cos\theta=\pm1$ and $V_{\text{axial}}=\pm(1/4\pi\epsilon_0)(p/r^{2})$. On the equatorial plane ($\theta=\pi/2$), $\cos\theta=0$ and $V_{\text{eq}}=0$ — every point on the perpendicular bisector of the dipole is at zero potential. Note that the dipole potential falls off as $1/r^{2}$, faster than the $1/r$ fall-off of a single point charge, because the contributions of the two charges nearly cancel. Q32. Derive an expression for the electrostatic potential energy of a system of two and three point charges, and use it to obtain the energy of a dipole in an external field. Answer: The work done in assembling charges from infinity (where they are taken to be at zero potential) is the potential energy of the system. Two charges. Bring $q_1$ first — no work done. Bring $q_2$ from infinity to a point at distance $r_{12}$ from $q_1$: $$U_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{12}}.$$ Three charges. Add $q_3$, which interacts with both $q_1$ and $q_2$: $$U=\frac{1}{4\pi\epsilon_0}\!\left[\frac{q_1 q_2}{r_{12}}+\frac{q_1 q_3}{r_{13}}+\frac{q_2 q_3}{r_{23}}\right].$$ The general result is $U=\dfrac{1}{4\pi\epsilon_0}\sum_{i Dipole in a uniform external field. If a dipole of moment $\vec p$ makes angle $\theta$ with a uniform field $\vec E$, the torque is $\vec\tau=\vec p\times\vec E$ and the work done in rotating from $\theta_0$ to $\theta$ is $$U(\theta)=-pE\cos\theta=-\vec p\cdot\vec E,$$ taking $U=0$ at $\theta=\pi/2$. The dipole has minimum energy when aligned with the field ($\theta=0$, $U=-pE$) and maximum energy when anti-parallel ($\theta=\pi$, $U=+pE$). Q33. Show that the relation $\vec E = -\nabla V$ correctly recovers the field of a point charge, and use it to obtain the radial field of a uniformly charged sphere. Answer: For a point charge, $V(r)=kq/r$ where $k=1/(4\pi\epsilon_0)$. By the chain rule $$E_r=-\frac{dV}{dr}=-\frac{d}{dr}\!\left(\frac{kq}{r}\right)=\frac{kq}{r^{2}},$$ which is Coulomb’s field directed radially outward for $q>0$, recovering Chapter 1. For a uniformly charged sphere of radius $R$ and total charge $Q$, the potential outside is the same as that of a point charge at the centre, $V(r)=kQ/r$ for $r\ge R$, so $E(r)=kQ/r^{2}$. Inside the sphere ($r Q34. A capacitor of capacitance $C$ is charged to a potential difference $V$ by a battery and then disconnected. A dielectric slab of constant $K$ that just fills the gap is now inserted. Compare $C$, $V$, $Q$, $E$ and stored energy $U$ before and after. Answer: Because the battery is disconnected, the charge $Q$ on the plates is conserved. The new capacitance becomes $C’=KC$. Using $C=Q/V$: The stored energy decreases by a factor $K$; the lost energy is used in polarising the dielectric (the slab is sucked into the gap, doing work). Q35. Repeat Q34 for the case where the battery remains connected. Answer: Now the voltage $V$ is held fixed by the battery. With $C\to KC$: The stored energy increases $K$-fold; the extra energy and the extra charge come from the battery (which also does work in pulling the slab in). Q36. A point charge $q=2\,\mu\text{C}$ is placed at the origin. Calculate (i) the electric potential at $r=10\,\text{cm}$ and (ii) the work done in moving a $+1\,\mu\text{C}$ test charge from $r=10\,\text{cm}$ to $r=20\,\text{cm}$. Answer: (i) $V(0.1)=kq/r=(9\times10^{9})(2\times10^{-6})/0.1=1.8\times10^{5}\,\text{V}$. (ii) $V(0.2)=(9\times10^{9})(2\times10^{-6})/0.2=9.0\times10^{4}\,\text{V}$. Work done by external agent $=q_0(V_f-V_i)=(1\times10^{-6})(9\times10^{4}-1.8\times10^{5})=-9.0\times10^{-2}\,\text{J}$. The negative sign means the field does positive work; the external agent absorbs $0.09\,\text{J}$. Q37. An $\alpha$-particle is accelerated through a potential difference of $1000\,\text{V}$. Find the kinetic energy gained, in joule and in eV. Answer: The $\alpha$-particle has charge $+2e$. KE gained $=qV=(2\times1.6\times10^{-19})(1000)=3.2\times10^{-16}\,\text{J}=2000\,\text{eV}=2\,\text{keV}$. Q38. Calculate the potential energy of a system of three charges $q_1=q_2=q_3=2\,\mu\text{C}$ placed at the vertices of an equilateral triangle of side $0.1\,\text{m}$. Answer: All three pairs are at equal distance $r=0.1\,\text{m}$: $$U=\frac{3kq^{2}}{r}=\frac{3\times(9\times10^{9})\times(2\times10^{-6})^{2}}{0.1}=1.08\,\text{J}.$$ Q39. Two charges $q_1=+1\,\mu\text{C}$ and $q_2=-1\,\mu\text{C}$ are placed $5\,\text{cm}$ apart, forming a dipole. Find (i) the dipole moment, (ii) the potential at $20\,\text{cm}$ on the axial line, and (iii) on the equatorial line. Answer: (i) $p=q\cdot 2a=(1\times10^{-6})(0.05)=5\times10^{-8}\,\text{C m}$. (ii) $V_{\text{axial}}=kp/r^{2}=(9\times10^{9})(5\times10^{-8})/(0.2)^{2}=1.125\times10^{4}\,\text{V}$. (iii) $V_{\text{eq}}=0$. Q40. A capacitor of $20\,\mu\text{F}$ is charged to $500\,\text{V}$. It is then connected in parallel with an uncharged $5\,\mu\text{F}$ capacitor. Find the common voltage and the energy lost. Answer: Total charge before $=Q=CV=(20\times10^{-6})(500)=10^{-2}\,\text{C}$. Total capacitance after parallel connection $=25\,\mu\text{F}$. Common voltage $V’=Q/C_{\text{tot}}=10^{-2}/(25\times10^{-6})=400\,\text{V}$. Initial energy $U_i=\tfrac12 CV^{2}=2.5\,\text{J}$; final energy $U_f=\tfrac12(25\times10^{-6})(400)^{2}=2.0\,\text{J}$. Energy lost $\Delta U=0.5\,\text{J}$. Q41. Find the equivalent capacitance between $A$ and $B$ when four capacitors of $4\,\mu\text{F}$ each are arranged in a Wheatstone-like square with one in each arm and a fifth $2\,\mu\text{F}$ in the diagonal. (Assume balanced bridge.) Answer: When opposite arms are equal ($4\,\mu\text{F}$ each), the bridge is balanced and the diagonal carries no charge — it can be removed. Then two arms in series give $4/2=2\,\mu\text{F}$, and the two such branches in parallel give $2+2=\mathbf{4\,\mu\text{F}}$. Q42. A parallel-plate capacitor with plate area $100\,\text{cm}^{2}$ and separation $1\,\text{mm}$ has air between its plates. It is charged to $200\,\text{V}$. Find the energy density of the field in the gap. Answer: $E=V/d=200/10^{-3}=2\times10^{5}\,\text{V m}^{-1}$. Energy density $u=\tfrac12\epsilon_0 E^{2}=\tfrac12(8.85\times10^{-12})(2\times10^{5})^{2}\approx 0.177\,\text{J m}^{-3}$. Q43. The plates of a parallel-plate capacitor are pulled apart slowly while it remains connected to a battery. Does the energy stored increase or decrease? Justify. Answer: $C=\epsilon_0 A/d$ decreases as $d$ increases, while $V$ stays fixed. $U=\tfrac12 CV^{2}$ therefore decreases. The battery absorbs the difference (charge flows back) and external work is done against the attraction of the plates, so the net energy of the system (battery + capacitor) is consistent. Q44. A spherical drop of mercury of radius $R$ has potential $V$. If $n$ such identical drops coalesce into a single drop, find the potential of the bigger drop. Answer: Volume conservation: $\tfrac43\pi R’^{3}=n\cdot\tfrac43\pi R^{3}$, so $R’=n^{1/3}R$. Total charge $Q’=nq$ where $q=4\pi\epsilon_0 RV$. New potential $V’=kQ’/R’=k(nq)/(n^{1/3}R)=n^{2/3}V$. Q45. A capacitor of capacitance $5\,\mu\text{F}$ is connected to a $20\,\text{V}$ supply. Calculate the magnitude of charge on each plate and the energy stored. Answer: $Q=CV=(5\times10^{-6})(20)=1.0\times10^{-4}\,\text{C}=100\,\mu\text{C}$. $U=\tfrac12 CV^{2}=\tfrac12(5\times10^{-6})(400)=1.0\times10^{-3}\,\text{J}=1\,\text{mJ}$. Q46. The electric potential at a point on the equatorial line of a short electric dipole is zero. Show that the electric field at that point is parallel to the dipole moment but oppositely directed. Answer: Although $V=0$ on the equatorial plane, $\vec E\ne 0$ because the equatorial point is closer to one charge than to the other in the limit of finite separation. Vector addition of the fields from $+q$ and $-q$ shows that the components perpendicular to the axis cancel, while the components parallel add: the resultant is $$\vec E_{\text{eq}}=-\frac{1}{4\pi\epsilon_0}\frac{\vec p}{r^{3}}\quad(r\gg a),$$ i.e. anti-parallel to $\vec p$, with magnitude $kp/r^{3}$ — half the axial field at the same distance. Q47. Why is the electric potential inside a charged conductor constant? Answer: Because $\vec E=0$ everywhere inside in equilibrium; therefore $-\nabla V=0$ and $V=$ const throughout the bulk and on the surface. Q48. Two metallic spheres of radii $R_1$ and $R_2$ are connected by a thin wire. Show that surface charge densities are inversely proportional to the radii. Answer: When connected, both spheres reach a common potential $V=kQ_1/R_1=kQ_2/R_2$, so $Q_1/R_1=Q_2/R_2$. Surface charge densities: $\sigma_i=Q_i/(4\pi R_i^{2})$. Hence $\sigma_1/\sigma_2=(Q_1/Q_2)(R_2/R_1)^{2}=(R_1/R_2)(R_2/R_1)^{2}=R_2/R_1$, i.e. $\sigma\propto 1/R$. This is why charge density (and thus the field) is highest at sharp points of a conductor — the principle of the lightning rod. Q49. Why is the capacitance of an isolated conductor smaller than that of a similar conductor placed near another grounded conductor? Answer: The grounded conductor lowers the potential on the charged conductor (induced opposite charges reduce $V$), so $C=Q/V$ is larger for given $Q$. This is the principle behind every practical capacitor — pair the conductor with a second one nearby. Q50. Distinguish between polar and non-polar dielectrics with one example each. Answer: A non-polar dielectric has molecules whose centres of positive and negative charges coincide in the absence of an external field, so the molecule has no permanent dipole moment (e.g. CO$_2$, H$_2$, O$_2$, methane). A polar dielectric has molecules with a permanent dipole moment due to asymmetric charge distribution (e.g. H$_2$O, HCl, NH$_3$). In both cases an external field induces (or aligns) molecular dipoles, producing net polarisation. Q51. Define dielectric strength. Why does it set a practical upper bound on the voltage of a capacitor? Answer: Dielectric strength is the maximum field a dielectric can sustain without breaking down (becoming conducting). For air it is $\approx 3\,\text{MV m}^{-1}$. If the operating field $E=V/d$ exceeds this value the dielectric ionises, the capacitor short-circuits and is destroyed; manufacturers therefore mark a maximum safe voltage. Q52. A hollow metal sphere of radius $R$ is given a charge $Q$. Sketch how the potential $V$ varies with distance $r$ from its centre. Answer: Inside the cavity ($0\le r\le R$) the field is zero so $V$ is constant and equal to its surface value $V_R=kQ/R$. Outside ($r>R$) the field is that of a point charge so $V$ falls off as $kQ/r$. The graph is a horizontal line at $kQ/R$ for $r\le R$, joined smoothly at $r=R$ to a $1/r$ decay for $r>R$. Q53. The SI unit of electric potential is Answer: (b) joule per coulomb (= volt). Q54. Three capacitors of $C$ each are connected in series. The equivalent capacitance is Answer: (b) $C/3$. Q55. The potential due to a point charge $q$ at distance $r$ varies as Answer: (b) $1/r$. Q56. If a dielectric slab of constant $K$ completely fills the gap of a parallel-plate capacitor of capacitance $C_0$, the new capacitance is Answer: (b) $KC_0$. Q57. Energy density in an electric field $E$ in vacuum is Answer: (c) $\tfrac12\epsilon_0 E^{2}$. Q58. Two metal spheres of radii $r_1 Answer: (c) $V_1>V_2$ (smaller sphere → higher potential since $V=kq/r$). Q59. Inside a hollow charged conductor the electric field is Answer: (c) zero. Q60. Work done in moving a charge $q$ from one point to another on the same equipotential surface is Answer: (d) $0$. Q61. A charge of $4\,\mu\text{C}$ is placed at the origin. Find the potential at $(3\,\text{cm},4\,\text{cm})$. Answer: $r=\sqrt{3^{2}+4^{2}}=5\,\text{cm}=0.05\,\text{m}$. $V=kq/r=(9\times10^{9})(4\times10^{-6})/0.05=7.2\times10^{5}\,\text{V}$. Q62. An electron of kinetic energy $200\,\text{eV}$ enters a region of zero potential. What is the change in its kinetic energy when it has moved through a potential difference of $-50\,\text{V}$? Answer: Because the electron has charge $-e$, moving through $\Delta V=-50\,\text{V}$ gives $\Delta U=q\Delta V=(-e)(-50)=+50\,\text{eV}$. By energy conservation, KE decreases by $50\,\text{eV}$, leaving $150\,\text{eV}$. Q63. A parallel-plate capacitor of capacitance $10\,\mu\text{F}$ is charged to $200\,\text{V}$. The battery is disconnected and a slab of dielectric constant $4$ is introduced. What is the new (i) capacitance, (ii) charge, (iii) voltage, (iv) energy stored? Answer: Initial $Q=CV=(10\times10^{-6})(200)=2\times10^{-3}\,\text{C}$; initial $U=\tfrac12 CV^{2}=0.2\,\text{J}$. After insertion: (i) $C’=KC=40\,\mu\text{F}$. (ii) $Q$ unchanged $=2\times10^{-3}\,\text{C}$ (battery disconnected). (iii) $V’=Q/C’=2\times10^{-3}/(40\times10^{-6})=50\,\text{V}$. (iv) $U’=\tfrac12 C’ V’^{2}=\tfrac12(40\times10^{-6})(2500)=0.05\,\text{J}$. Note $U’/U=1/4=1/K$, as expected. Q64. A charge of $10\,\mu\text{C}$ is uniformly distributed on a metal sphere of radius $10\,\text{cm}$. Calculate the potential at (i) the surface, (ii) the centre, (iii) a point $5\,\text{cm}$ from the centre. Answer: (i) $V_R=kq/R=(9\times10^{9})(10\times10^{-6})/0.1=9\times10^{5}\,\text{V}$. (ii) For a conductor $V$ is constant inside, so $V_{centre}=9\times10^{5}\,\text{V}$. (iii) Same — the field is zero inside, $V_{5\text{cm}}=9\times10^{5}\,\text{V}$. Q65. Two capacitors $4\,\mu\text{F}$ and $6\,\mu\text{F}$ are connected in series across a $200\,\text{V}$ battery. Find the charge on each and the voltage across each. Answer: $C_s=(4\times6)/(4+6)=2.4\,\mu\text{F}$. Charge on each (same in series) $=C_s V=(2.4\times10^{-6})(200)=4.8\times10^{-4}\,\text{C}$. Voltages $V_1=Q/C_1=(4.8\times10^{-4})/(4\times10^{-6})=120\,\text{V}$; $V_2=Q/C_2=80\,\text{V}$. Check: $V_1+V_2=200\,\text{V}$. Q66. A capacitor of $100\,\mu\text{F}$ is charged to $50\,\text{V}$. Find the energy stored. If half the charge is suddenly removed (e.g. by sharing with an identical uncharged capacitor — but for this part assume $Q$ becomes $Q/2$), what is the new stored energy? Answer: $U_1=\tfrac12 CV^{2}=\tfrac12(100\times10^{-6})(2500)=0.125\,\text{J}=125\,\text{mJ}$. With charge halved, $V’=V/2=25\,\text{V}$ on the same $C$, so $U_2=\tfrac12 C V’^{2}=\tfrac12 (100\times10^{-6})(625)=0.03125\,\text{J}\approx 31.25\,\text{mJ}$. Ratio $U_2/U_1=1/4$. Q67. A potential difference of $1000\,\text{V}$ is applied across a $4\,\mu\text{F}$ capacitor. Calculate (i) the charge on each plate, (ii) the energy stored, (iii) the energy stored if the capacitor is connected in series with another $4\,\mu\text{F}$ across the same battery. Answer: (i) $Q=CV=(4\times10^{-6})(1000)=4\,\text{mC}$. (ii) $U=\tfrac12 CV^{2}=\tfrac12(4\times10^{-6})(10^{6})=2\,\text{J}$. (iii) Series gives $C_s=2\,\mu\text{F}$, so $U’=\tfrac12 C_s V^{2}=1\,\text{J}$. Q68. The plates of a parallel-plate capacitor have area $A=400\,\text{cm}^{2}$ and are separated by $d=1\,\text{mm}$. A potential difference of $200\,\text{V}$ is applied. Find (i) the capacitance, (ii) the charge, (iii) the field, (iv) the energy density. Answer: (i) $C=\epsilon_0 A/d=(8.85\times10^{-12})(0.04)/(10^{-3})=3.54\times10^{-10}\,\text{F}=354\,\text{pF}$. (ii) $Q=CV=(354\times10^{-12})(200)=7.08\times10^{-8}\,\text{C}\approx 70.8\,\text{nC}$. (iii) $E=V/d=2\times10^{5}\,\text{V m}^{-1}$. (iv) $u=\tfrac12 \epsilon_0 E^{2}=\tfrac12(8.85\times10^{-12})(4\times10^{10})\approx 0.177\,\text{J m}^{-3}$. Q69. A point charge $q=+1\,\mu\text{C}$ is at the centre of a cube of side $0.5\,\text{m}$. (i) What is the potential at any face centre? (ii) What is the work done in moving a $-1\,\mu\text{C}$ test charge from one face centre to the opposite face centre? Answer: (i) Distance from centre to face centre $=0.25\,\text{m}$. $V=kq/r=(9\times10^{9})(10^{-6})/0.25=3.6\times10^{4}\,\text{V}$. (ii) Both face centres are equidistant from the cube centre, so they are at the same potential and the work done $=q_0(V_f-V_i)=0$. Q70. A network has $4\,\mu\text{F}$ and $6\,\mu\text{F}$ in parallel, and this combination is in series with $5\,\mu\text{F}$ across a $100\,\text{V}$ supply. Find the equivalent capacitance and the charge supplied by the battery. Answer: Parallel: $4+6=10\,\mu\text{F}$. Series with $5\,\mu\text{F}$: $C_{eq}=(10\times5)/(10+5)=10/3\,\mu\text{F}\approx3.33\,\mu\text{F}$. Charge from battery $=C_{eq} V=(10/3)\times10^{-6}\times100\approx 3.33\times10^{-4}\,\text{C}\approx 333\,\mu\text{C}$. Q71. The energy stored in a $5\,\mu\text{F}$ capacitor charged to $100\,\text{V}$ is shared with an uncharged $20\,\mu\text{F}$ capacitor. Find the common voltage and the percentage energy lost. Answer: Initial charge $Q=(5\times10^{-6})(100)=5\times10^{-4}\,\text{C}$. After sharing across $25\,\mu\text{F}$, $V’=Q/C_{tot}=(5\times10^{-4})/(25\times10^{-6})=20\,\text{V}$. $U_i=\tfrac12 (5\times10^{-6})(10^{4})=0.025\,\text{J}$. $U_f=\tfrac12(25\times10^{-6})(400)=0.005\,\text{J}$. Loss fraction $=(0.025-0.005)/0.025=0.8=80\%$. Q72. An air-filled parallel-plate capacitor has capacitance $C_0$. A dielectric of constant $K$ and thickness $d/2$ is inserted, parallel to the plates. Find the new capacitance. Answer: Use the partial-dielectric formula with $t=d/2$: $$C=\frac{\epsilon_0 A}{d-\dfrac{d}{2}\!\left(1-\dfrac{1}{K}\right)}=\frac{2K}{K+1}\,C_0.$$ For $K=2$, $C=4C_0/3$; for $K\to\infty$, $C\to 2C_0$ (gap effectively halved by the metallic limit). This completes the ASSEB Class 12 Physics Chapter 2 — Electrostatic Potential and Capacitance question-answer set in English Medium. Revise the formulas, redraw the SVG diagrams from memory, and practise the numericals once without looking; you will be ready for any board-style question on this chapter. For more chapters in the ASSEB Higher Secondary syllabus, browse the rest of HSLC Guru.Key Formulas at a Glance
Quantity Formula SI unit Electric potential of point charge $V = \dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r}$ volt (V) Potential difference $V_A – V_B = -\int_B^A \vec{E}\cdot d\vec{l}$ V Field from potential $E = -\dfrac{dV}{dr}$ V m$^{-1}$ Potential of dipole (axial) $V = \dfrac{1}{4\pi\epsilon_0}\dfrac{p}{r^{2}}$ V Potential of dipole (equatorial) $V = 0$ V Potential energy of two charges $U = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r_{12}}$ joule (J) Capacitance $C = Q/V$ farad (F) Parallel-plate (vacuum) $C = \epsilon_0 A/d$ F Parallel-plate (dielectric) $C = K\epsilon_0 A/d$ F Series combination $\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots$ F Parallel combination $C = C_1 + C_2 + \cdots$ F Energy stored $U = \tfrac{1}{2}CV^{2}=\dfrac{Q^{2}}{2C}=\tfrac{1}{2}QV$ J Energy density $u = \tfrac{1}{2}\epsilon_0 E^{2}$ J m$^{-3}$ Diagrams
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Glossary
Term Meaning Electric potential Work done per unit positive test charge to bring it from infinity to a point. Potential difference $V_A-V_B$; work per unit charge to move from $B$ to $A$. Equipotential surface Surface on which $V$ is constant; $\vec E$ is normal to it. Dipole moment $\vec p = q(2\vec a)$, directed from $-q$ to $+q$. Conductor Material with free charges; $\vec E=0$ inside in equilibrium. Dielectric Insulator that polarises in an external field. Dielectric constant $K$ Ratio $C/C_0$; also the factor by which the field is reduced inside the slab. Capacitance $C$ $Q/V$; charge per unit voltage stored in a capacitor. Farad (F) SI unit of capacitance, $1\,\text{F}=1\,\text{C/V}$. Energy density $u$ $\tfrac{1}{2}\epsilon_0 E^{2}$; electrostatic energy per unit volume. Theory Notes — Detailed
Long-Answer Derivations
Quantity Before slab After slab Ratio after/before Charge $Q$ $Q$ $Q$ $1$ Capacitance $C$ $KC$ $K$ Voltage $V$ $V/K$ $1/K$ Field $E$ between plates $E_0$ $E_0/K$ $1/K$ Energy $U=Q^{2}/2C$ $Q^{2}/2C$ $Q^{2}/2KC$ $1/K$ Quantity Before slab After slab Ratio after/before Voltage $V$ $V$ $V$ $1$ Capacitance $C$ $KC$ $K$ Charge $Q=CV$ $CV$ $KCV$ $K$ Field $E=V/d$ $E_0$ $E_0$ $1$ Energy $U=\tfrac12 CV^{2}$ $\tfrac12 CV^{2}$ $\tfrac12 KCV^{2}$ $K$ More Numerical Problems
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Practice Set — Mixed Numericals
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Glossary (Extended)
Term Meaning Electronvolt (eV) Energy gained by an electron passing through $1\,\text{V}$ p.d.; $1\,\text{eV}=1.6\times10^{-19}\,\text{J}$. Polarisation $\vec P$ Dipole moment per unit volume of a dielectric. Surface charge density $\sigma$ Charge per unit area; SI unit $\text{C m}^{-2}$. Linear charge density $\lambda$ Charge per unit length; SI unit $\text{C m}^{-1}$. Volume charge density $\rho$ Charge per unit volume; SI unit $\text{C m}^{-3}$. Dielectric strength Maximum field the dielectric can withstand without breakdown. Permittivity $\epsilon$ $\epsilon = K\epsilon_0$; characterises the medium. Microfarad ($\mu$F) $10^{-6}\,\text{F}$ — common practical unit. Picofarad (pF) $10^{-12}\,\text{F}$ — used in high-frequency and small-signal circuits. Electrostatic shielding Use of a conductor’s interior (where $E=0$) to protect sensitive equipment.