Electrostatic Potential And Capacitance
Chapter 2
ELECTROSTATIC POTENTIAL AND CAPACITANCE
MCQ I
2.1 A capacitor of $4 \mu F$ is connected as shown in the circuit (Fig. 2.1). The internal resistance of the battery is $0.5 \Omega$. The amount of charge on the capacitor plates will be
(a) 0
(b) $4 \mu C$
(c) $16 \mu C$
(d) $8 \mu C$
Fig. 2.1
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Answer: (d)2.2 A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge
(a) remains a constant because the electric field is uniform.
(b) increases because the charge moves along the electric field.
(c) decreases because the charge moves along the electric field.
(d) decreases because the charge moves opposite to the electric field.
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Answer: (c)2.3 Figure 2.2 shows some equipotential lines distributed in space. A charged object is moved from point $A$ to point $B$.
(i)
(ii)
Fig. 2.2
(iii)
(a) The work done in Fig. (i) is the greatest.
(b) The work done in Fig. (ii) is least.
(c) The work done is the same in Fig. (i), Fig. (ii) and Fig. (iii).
(d) The work done in Fig. (iii) is greater than Fig. (ii)but equal to that in Fig. (i).
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Answer: (c)2.4 The electrostatic potential on the surface of a charged conducting sphere is $100 V$. Two statments are made in this regard:
$S_1$ : At any point inside the sphere, electric intensity is zero.
$S_2$ : At any point inside the sphere, the electrostatic potential is $100 V$.
Which of the following is a correct statement?
(a) $S_1$ is true but $S_2$ is false.
(b) Both $S_1 & S_2$ are false.
(c) $S_1$ is true, $S_2$ is also true and $S_1$ is the cause of $S_2$.
(d) $S_1$ is true, $S_2$ is also true but the statements are independant.
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Answer: (c)2.5 Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately
(a) spheres.
(b) planes.
(c) paraboloids
(d) ellipsoids.
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Answer: (a)2.6 A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_1$ and dielectric constant $k_1$ and
Fig. 2.3
the other has thickness $d_2$ and dielectric constant $k_2$ as shown in Fig. 2.3. This arrangement can be thought as a dielectric slab of thickness $d(=d_1+d_2)$ and effective dielectric constant $k$. The $k$ is
(a) $\frac{k_1 d_1+k_2 d_2}{d_1+d_2}$
(b) $\frac{k_1 d_1+k_2 d_2}{k_1+k_2}$
(c) $\frac{k_1 k_2(d_1+d_2)}{(k_1 d_1+k_2 d_2)}$
(d) $\frac{2 k_1 k_2}{k_1+k_2}$
MCQ II
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Answer: (c)2.7 Consider a uniform electric field in the $\hat{\mathbf{z}}$ direction. The potential is a constant
(a) in all space.
(b) for any $x$ for a given $z$.
(c) for any $y$ for a given $z$.
(d) on the $x$ - $y$ plane for a given $z$.
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Answer: (b), (c), (d)2.8 Equipotential surfaces
(a) are closer in regions of large electric fields compared to regions of lower electric fields.
(b) will be more crowded near sharp edges of a conductor.
(c) will be more crowded near regions of large charge densities.
(d) will always be equally spaced.
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Answer: (a), (b), (c)2.9 The work done to move a charge along an equipotential from A to B
(a) cannot be defined as $-\int_A^{B} \mathbf{E} . d \mathbf{1}$
(b) must be defined as $-\int_A^{B} \mathbf{E} \cdot \boldsymbol{{}d} \mathbf{1}$
(c) is zero.
(d) can have a non-zero value.
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Answer: (b), (c)2.10 In a region of constant potential
(a) the electric field is uniform
(b) the electric field is zero
(c) there can be no charge inside the region.
(d) the electric field shall necessarily change if a charge is placed outside the region.
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Answer: (b), (c)2.11 In the circuit shown in Fig. 2.4. initially key $K_1$ is closed and key $K_2$ is open. Then $K_1$ is opened and $K_2$ is closed (order is important). [Take $Q_1{ }^{\prime}$ and $Q_2{ }^{\prime}$ as charges on $C_1$ and $C_2$ and $V_1$ and $V_2$ as voltage respectively.]
Then
Fig. 2.4
(a) charge on $C_1$ gets redistributed such that $V_1=V_2$
(b) charge on $C_1$ gets redistributed such that $Q_1{ }^{\prime}=Q_2{ }^{\prime}$
(c) charge on $C_1$ gets redistributed such that $C_1 V_1+C_2 V_2=C_1 E$
(d) charge on $C_1$ gets redistributed such that $Q_1{ }^{\prime}+Q_2{ }^{\prime}=Q$
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Answer: (a), (d)2.12 If a conductor has a potential $V \neq 0$ and there are no charges anywhere else outside, then
(a) there must be charges on the surface or inside itself.
(b) there cannot be any charge in the body of the conductor.
(c) there must be charges only on the surface.
(d) there must be charges inside the surface.
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Answer: (a), (b)2.13 A parallel plate capacitor is connected to a battery as shown in Fig. 2.5. Consider two situations:
A: Key $K$ is kept closed and plates of capacitors are moved apart using insulating handle.
B: Key $K$ is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
(a) In A : $Q$ remains same but $C$ changes.
(b) In B : $V$ remains same but $C$ changes.
(c) In A : $V$ remains same and hence $Q$ changes.
Fig. 2.5
(d) In B : $Q$ remains same and hence $V$ changes. smaller sphere is more or less than that of the larger one.
VSA
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Answer: (c) and (d)2.14 Consider two conducting spheres of radii R1 and R2 with R1 > R2. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.
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Answer: More.2.15 Do free electrons travel to region of higher potential or lower potential?
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Answer: Higher potential.2.16 Can there be a potential difference between two adjacent conductors carrying the same charge?
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Answer: Yes, if the sizes are different.2.17 Can the potential function have a maximum or minimum in free space?
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Answer: No.2.18 A test charge q is made to move in the electric field of a point charge Q along two different closed paths (Fig. 2.6). First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?
Fig. 2.6
SA
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Answer: As electric field is conservative, work done will be zero in both the cases.2.19 Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.
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Answer: Suppose this were not true. The potential just inside the surface would be different from that at the surface resulting in a potential gradient. This would mean that there are field lines pointing inwards or outwards from the surface. These lines cannot at the other end be again on the surface, since the surface is equipotential. Thus, this is possible only if the other end of the lines are at charges inside, contradicting the premise. Hence, the entire volume inside must be at the same potential.2.20 A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.
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Answer: C will decrease Energy stored $=\frac{1}{2} C V^{2}$ and hence will increase. Electric field will increase.
Charge stored will remain the same.
$V$ will increase.
2.21 Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.
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Answer: Consider any path from the charged conductor to the uncharged conductor along the electric field. The potential will continually decrease along this path. A second path from the uncharged conductor to infinity will again continually lower the potential further. Hence this result.
2.22 Calculate potential energy of a point charge $-q$ placed along the axis due to a charge $+Q$ uniformly distributed along a ring of radius $R$. Sketch P.E. as a function of axial distance $z$ from the centre of the ring. Looking at graph, can you see what would happen if $-q$ is displaced slightly from the centre of the ring (along the axis)?
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Answer: $U=\frac{-q Q}{4 \pi \varepsilon_0 R \sqrt{1+z^{2} / R^{2}}}$
The variation of potential energy with $z$ is shown in the figure.
The charge $-q$ displaced would perform oscillations. We cannot conclude anything just by looking at the graph.
2.23 Calculate potential on the axis of a ring due to charge $Q$ uniformly distributed along the ring of radius $R$.
LA
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Answer: $ \quad V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\sqrt{R^{2}+z^{2}}}$2.24 Find the equation of the equipotentials for an infinite cylinder of radius $r_0$, carrying charge of linear density $\lambda$.
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Answer: To find the potential at distance $r$ from the line consider the electric field. We note that from symmetry the field lines must be radially outward. Draw a cylindrical Gaussian surface of radius $r$ and length $l$. Then
$\oint \mathbf{E} . d \mathbf{S}=\frac{1}{\varepsilon_0} \lambda 1$
Or $E_r 2 \pi rl=\frac{1}{\varepsilon_0} \lambda l$
$\Rightarrow E_r=\frac{\lambda}{2 \pi \varepsilon_0 r}$
Hence, if $r_0$ is the radius,
$ V(r)-V(r_ 0)= - \int_{r_0^r}^{r} \mathbf{E} \cdot d \mathbf{l}=\frac{\lambda} {2 \pi \varepsilon_ 0} \ln \frac{r _0}{r}$
For a given $V$,
$\ln \frac{r}{r_0}=-\frac{2 \pi \varepsilon_0}{\lambda}[V(r)-V(r_0)]$
$\Rightarrow r=r_0 e^{-2 \pi \varepsilon_0 V r_0 / \lambda} \cdot e^{+2 \pi \varepsilon_0 V(r) / \lambda}$
The equipotential surfaces are cylinders of radius
$r=r_0 e^{-2 \pi \varepsilon_0[V(r)-V(r_0)] / \lambda}$
2.25 Two point charges of magnitude $+q$ and $-q$ are placed at $(-d / 2,0,0)$ and $(d / 2,0,0)$, respectively. Find the equation of the equipoential surface where the potential is zero.
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Answer: Let the plane be at a distance $x$ from the origin. The potential at the point $P$ is
$\frac{1}{4 \pi \varepsilon_0} \frac{q}{[(x+d / 2)^{2}+h^{2}]^{1 / 2}}-\frac{1}{4 \pi \varepsilon_0} \frac{q}{[(x-d / 2)^{2}+h^{2}]^{1 / 2}}$
If this is to be zero.
$\frac{1}{[(x+d / 2)^{2}+h^{2}]^{1 / 2}}=\frac{1}{[(x-d / 2)^{2}+h^{2}]^{1 / 2}}$
Or, $(x-d / 2)^{2}+h^{2}=(x+d / 2)^{2}+h^{2}$
$\Rightarrow x^{2}-d x+d^{2} / 4=x^{2}+d x+d^{2} / 4$
Or, $2 d x=0$
$\Rightarrow x=0$
The equation is that of a plane $x=0$.
2.26 A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $(U)$ as $\varepsilon=\alpha U$ where $\alpha=2 V^{-1}$.A similar capacitor with no dielectric is charged to $U_0=78 V$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
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Answer: Let the final voltage be $U$ : If $C$ is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is
$Q_1=C U$
The capacitor with the dielectric has a capacitance $\varepsilon C$. Hence the charge on the capacitor is
$Q_2=\varepsilon U=\alpha C U^{2}$
The initial charge on the capacitor that was charged is
$Q_0=CU_0$
From the conservation of charges,
$B_0=Q_1+Q_2$
Or, $CU_0=CU+\alpha CU^{2}$
$\Rightarrow \alpha U^{2}+U-u_0=0$
$\therefore U=\frac{-1 \pm \sqrt{1+4 \alpha U_0}}{2 \alpha}$
$ \begin{aligned} & =\frac{-1 \pm \sqrt{1+624}}{4} \\ & =\frac{-1 \pm \sqrt{625}}{4} \text{ volts } \end{aligned} $
As $U$ is positive
$ U=\frac{\sqrt{625}-1}{4}=\frac{24}{4}=6 V $
2.27 A capacitor is made of two circular plates of radius $R$ each, separated by a distance $d«R$. The capacitor is connected to a constant voltage. A thin conducting disc of radius $r«R$ and thickness $t«r$ is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is $m$.
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Answer: When the disc is in touch with the bottom plate, the entire plate is a equipotential. A change $q^{\prime}$ is transferred to the disc.
The electric field on the disc is
$=\frac{V}{d}$
$\therefore q^{\prime}=-\varepsilon_0 \frac{V}{d} \pi r^{2}$
The force acting on the disc is
$ -\frac{V}{d} \times q^{\prime}=\varepsilon_0 \frac{V^{2}}{d^{2}} \pi r^{2} $
If the disc is to be lifted, then
$ \begin{aligned} & \varepsilon_0 \frac{V^{2}}{d^{2}} \pi r^{2}=m g \\ & \Rightarrow V=\sqrt{\frac{m g d^{2}}{\pi \varepsilon_0 r^{2}}} \end{aligned} $
2.28 (a) In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3) e] and two down quarks [charges $-(1 / 3)$ e]. Assume that they have a triangle configuration with side length of the order of $10^{-15} m$. Calculate electrostatic potential energy of neutron and compare it with its mass 939 $MeV$.
(b) Repeat above exercise for a proton which is made of two up and one down quark.
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Answer: $ U = \frac{1}{4 \pi \varepsilon _0}$ $\begin{Bmatrix} \frac{q_d q_d}{r} - \frac{q_u q_d}{r} - \frac{q_u q_d} {r} \end{Bmatrix}$
$=\frac{9 \times 10^{9}}{10^{-15}}(1.6 \times 10^{-19})^{2}{(1 / 3)^{2}-(2 / 3)(1 / 3)-(2 / 3)(1 / 3)}$
$=2.304 \times 10^{-13}{\frac{1}{9}-\frac{4}{9}}=-7.68 \times 10^{-14} J$
$ =4.8 \times 10^{5} eV=0.48 MeV=5.11 \times 10^{-4}(m_n c^{2}) $
Before contact
$Q_1=\sigma .4 \pi R^{2}$
$Q_2=\sigma .4 \pi(2 R^{2})=4(\sigma .4 \pi R^{2})=4 Q_1$
After contact :
$ \begin{aligned} Q_1^{\prime}+Q_2^{\prime} & =Q_1+Q_2=5 Q_1 \\ & =5(\sigma .4 \pi R^{2}) \end{aligned} $
They will be at equal potentials:
$\frac{Q_1{ }^{\prime}}{R}=\frac{Q_2{ }^{\prime}}{2 R}$
$\therefore Q_2{ }^{\prime}=2 Q^{\prime}$.
$\therefore 3 Q_1{ }^{\prime}=5(\sigma .4 \pi R^{2})$
$\therefore Q_1{ }^{\prime}=\frac{5}{3}(\sigma .4 \pi R^{2})$ and $Q_2^{\prime}=\frac{10}{3}(\sigma \cdot 4 \pi R^{2})$
$\therefore \sigma_1=5 / 3 \sigma$ and $\therefore \sigma_2=\frac{5}{6} \sigma$.
2.29 Two metal spheres, one of radius $R$ and the other of radius $2 R$, both have same surface charge density $\sigma$. They are brought in contact and separated. What will be new surface charge densities on them?
2.30 In the circuit shown in Fig. 2.7, initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitors.
Then $K_1$ was opened and $K_2$ was closed (order is important), What will be the charge on each capacitor now? $[C=1 \mu F]$
Fig. 2.7
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Answer: Initially : $V \propto \frac{1}{C}$ and $V_1+V_2=E$
$\Rightarrow V_1=3 V$ and $V_2=6 V$
$\therefore Q_1=C_1 V_1=6 C \times 3=18 \mu C$
$Q_2=9 \mu C$ and $Q_3=0$
Later : $Q_2=Q_2^{\prime}+Q_3$
with $C_2 V+C_3 V=Q_2 \quad \Rightarrow V=\frac{Q_2}{C_2+C_3}=(3 / 2) V$
$Q_2{ }^{\prime}=(9 / 2) \mu C$ and $Q_3{ }^{\prime}=(9 / 2) \mu C$
2.31 Calculate potential on the axis of a disc of radius $R$ due to a charge $Q$ uniformly distributed on its surface.
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Answer: $ \quad \sigma=\frac{Q}{\pi R^{2}}$
$d U=\frac{1}{4 \pi \varepsilon_0} \frac{\sigma \cdot 2 \pi r d r}{\sqrt{r^{2}+z^{2}}}$
$\therefore U=\frac{\pi \sigma}{4 \pi \varepsilon_0} \int_0^{R} \frac{2 r d r}{\sqrt{r^{2}+z^{2}}}$
2.32 Two charges $q_1$ and $q_2$ are placed at $(0,0, d)$ and $(0,0,-d)$ respectively. Find locus of points where the potential a zero.
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Answer: $ \begin{aligned} & \quad=\frac{2 \pi \sigma}{4 \pi \varepsilon_0}[\sqrt{r^{2}+z^{2}}]_0^{R}=\frac{2 \pi \sigma}{4 \pi \varepsilon_0}[\sqrt{R^{2}+z^{2}}-z] \\ & =\frac{2 Q}{4 \pi \varepsilon_0 R^{2}}[\sqrt{R^{2}+z^{2}}-z] \\ & \frac{q_1}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}+\frac{q_2}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}}=0 \\ & \therefore \frac{q_1}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}=\frac{-q_2}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}} \end{aligned} $
Thus, to have total potential zero, $q_1$ and $q_2$ must have opposite signs. Squaring and simplifying, we get.
$x^{2}+y^{2}+z^{2}+[\frac{(q_1 / q_2)^{2}+1}{(q_1 / q_2)^{2}-1}](2 z d)+d^{2}=0$
This is the equation of a sphere with centre at $(0,0,-2 d[\frac{q_1{ }^{2}+q_1{ }^{2}}{q_1^{2}-q_1^{2}}])$.
Note : if $q_1=-q_2 \Rightarrow$ Then $z=0$, which is a plane through mid-point.
2.33 Two charges $-q$ each are separated by distance $2 d$. A third charge $+q$ is kept at mid point $O$. Find potential energy of $+q$ as a function of small distance $x$ from $O$ due to $-q$ charges. Sketch P.E. v/s $x$ and convince yourself that the charge at $O$ is in an unstable equilibrium.
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Answer: $U=\frac{1}{4 \pi \varepsilon_0}{\frac{-q^{2}}{(d-x)}+\frac{-q^{2}}{(d-x)}}$
$U=\frac{-q^{2}}{4 \pi \varepsilon_0} \frac{2 d}{(d^{2}-x^{2})}$
$\frac{dU}{dx}=\frac{-q^{2} \cdot 2 d}{4 \pi \in_0} \cdot \frac{2 x}{(d^{2}-x^{2})^{2}}$
$ \begin{aligned} & U_0=\frac{2 q^{2}}{4 \pi \varepsilon_0 d} \quad \frac{dU}{dx}=0 \text{ at } x=0 \\ & x=0 \text{ is an equilibrium point. } \\ & \frac{d^{2} U}{dx^{2}}=(\frac{-2 d q^{2}}{4 \pi \in_0})[\frac{2}{(d^{2}-x^{2})^{2}}-\frac{8 x^{2}}{(d^{2}-x^{2})^{3}}] \end{aligned} $
$ =(\frac{-2 d q^{2}}{4 \pi \in_0}) \frac{1}{(d^{2}-x^{2})^{3}}[2(d^{2}-x^{2})^{2}-8 x^{2}] $
At $x=0$
$\frac{d^{2} U}{dx^{2}}=(\frac{-2 d q^{2}}{4 \pi \epsilon_0})(\frac{1}{d^{6}})(2 d^{2})$, which is $<0$.
Hence, unstable equilibrium.
