# Electrostatic Fields

 Question 1
As shown in the figure below, two concentric conducting spherical shells, centered at $r=0$ and having radii $r=c$and $r=d$ are maintained at potentials such that the potential $V(r)$ at $r=c$ is $V_1$ and $V(r)$ at $r=d$ is $V_2$. Assume that $V(r)$ depends only on $r$, where $r$ is the radial distance. The expression for $V(r)$ in the region between $r=c$and $r=d$ is A $V(r)=\frac{cd(V_2-V_1)}{(d-c)r}-\frac{V_1c+V_2d-2V_1d}{d-c}$ B $V(r)=\frac{cd(V_1-V_2)}{(d-c)r}-\frac{V_2d-V_1c}{d-c}$ C $V(r)=\frac{cd(V_1-V_2)}{(d-c)r}-\frac{V_1c-V_2c}{d-c}$ D $V(r)=\frac{cd(V_2-V_1)}{(d-c)r}-\frac{V_2c-V_1c}{d-c}$
GATE EE 2022   Electromagnetic Theory
Question 1 Explanation:
We have, Laplace equation
$\triangledown ^2V=0$ ...(1)
where, $\triangledown ^2V=\frac{1}{r^2}\frac{\partial }{\partial r}\left ( r^2\frac{\partial V}{\partial r} \right )+\frac{1}{r^2 \sin \theta }\frac{\partial }{\partial \theta }\left ( \sin \theta \frac{\partial V}{\partial \theta } \right )+ \frac{1}{r^2 \sin ^2\theta } \frac{\partial^2 V}{\partial \phi ^2}$
V will have only radial component.
From equation (1)
$\frac{1}{r^2}\frac{\partial }{\partial r}\left ( r^2\frac{\partial V}{\partial r} \right )=0$
Integrate both side $r^2\frac{\partial V}{\partial r} =P$
Again integrate, $V=-\frac{P}{r}+Q$
Here, P and Q are constant.
Given: at $r=c, V=V_1$
$V_1=-\frac{P}{r}+Q$ ...(2)
at $r=d, V=V_2$
$V_2=-\frac{P}{d}+Q$ ...(3)
From eq. (2) & (3)
$P=\frac{(V_2-V_1)cd}{d-c}$
$Q=V_1+\frac{d(V_2-V_1)}{d-c}$
From eq. (1),
$V=-\frac{(V_2-V_1)cd}{r(d-c)}+V_1+\frac{(V_2-V_1)cd}{(d-c)}$
$V=\frac{(V_1-V_2)cd}{r(d-c)}+\frac{(V_2d-V_1c)}{(d-c)}$
 Question 2
Consider a large parallel plate capacitor. The gap 'd' between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of V volts and then disconnected from the source. If the dielectric slab is pulled out completely, then the ratio of the new electric field $E_{2}$ in the gap to the original electric field $E_{1}$ is ________.
 A 2 B 5 C 6 D 8
GATE EE 2021   Electromagnetic Theory
Question 2 Explanation: If voltage source is removed then in both cases charge O is constant.
\begin{aligned} \text { Case-1: }\left(Q_{1},=Q ; V_{1}=\right.&V) \\ Q_{1} &=C_{1} V_{1} \\ Q &=\frac{\epsilon_{0}(5) A}{d} V_{1} \\ Q &=5\left(\frac{\epsilon_{0} A}{d}\right) V_{1} \\ \text { Case-2: }\left(Q_{2}=Q ; V_{2}\right) & \\ Q_{2} &=C_{2} V_{2} \\ Q &=\frac{\epsilon_{0}(1) A}{d} V_{2} \\ Q &=\left(\frac{\epsilon_{0} A}{d}\right) V_{2} \end{aligned}
Equation (i) is equal to equation (ii)
\begin{aligned} \Rightarrow 5\left(\frac{\epsilon_{0} A}{d}\right) V_{1} &=\left(\frac{\epsilon_{0} A}{d}\right) V_{2} \\ \Rightarrow \qquad \qquad 5 V_{1} &=V_{2} \\ \frac{V_{2}}{V_{1}} &=5 \\ E_{1} &=\frac{V_{1}}{d} ; E_{2}=\frac{V_{2}}{d} \\ \frac{E_{2}}{E_{1}} &=\frac{V_{2} / d}{V_{1} / d} \\ \Rightarrow \qquad \qquad \frac{E_{2}}{E_{1}} &=\frac{V_{2}}{V_{1}} \end{aligned}
Put equation (iii) in equation (iv),
$\Rightarrow \Rightarrow \qquad \qquad \frac{E_{2}}{E_{1}}=5$
 Question 3
A $1\;\mu C$ point charge is held at the origin of a cartesian coordinate system. If a second point charge of $10\;\mu C$ is moved from $\left ( 0,10,0 \right )$ to $\left ( 5,5,5\right )$ and subsequently to $\left ( 5,0,0\right )$, then the total work done is ___________$\text{mJ}$. (Round off to 2 decimal places).
Take $\dfrac{1}{4\pi \varepsilon _{0}}=9\times 10^{9}$ in SI units. All coordinates are in meters.
 A 5.25 B 9 C 8.62 D 2.47
GATE EE 2021   Electromagnetic Theory
Question 3 Explanation: In this case work done is independent of type of path but depends an intial and final point.
$\begin{array}{ccccccccc} x & y & z & & r & \theta & \phi \\ (0, & 10, & 0) & \rightarrow & (10, & 90^{\circ}, & \left.90^{\circ}\right) & \text { Initial point }\\ x & y & z & & r & \theta & \phi & \\ (5, & 5, & 5) & \rightarrow & (5 \sqrt{3}, & 54.73^{\circ}, & \left.45^{\circ}\right) & \text { Intermediate point } \\ x & y & z & & r & \theta & \phi & \\ (5, & 0, & 0) & \rightarrow & (5, & 90^{\circ}, & \left.0^{\circ}\right) & \text { Final point } \end{array}$
Work done (by external source) in moving Q-charge ($10 \mu C$) in the presence of electric field $\vec{E}$ is
\begin{aligned} W&=-Q \int_{\text {initial point }}^{\text {final point }} \vec{E} \cdot d \vec{l}=-\left(10 \times 10^{-6}\right) \int_{r=10}^{r=5} \frac{9 \times 10^{3}}{r^{2}} \hat{a}_{r} \cdot d r \hat{a}_{r} \\ W&=-10 \times 10^{-6}\left(9 \times 10^{3}\right)\left(-\frac{1}{r}\right)_{r=10}^{5} \\ &=9 \times 10^{-2}\left[\frac{1}{5}-\frac{1}{10}\right]=9 \times 10^{-2}\left[\frac{10-5}{50}\right]=9 \times 10^{-2}\left(10^{-1}\right) \\ W&=9 \mathrm{~mJ} \end{aligned}
 Question 4
Let $a_r, a_\phi \; and \; a_z$ be unit vectors along $r, \phi \; and \; z$ directions, respectively in the cylindrical coordinate system. For the electric flux density given by $D = (a_r15 + a_\phi 2r - a_z3rz)$ Coulomb/$m^2$, the total electric flux, in Coulomb, emanating from the volume enclosed by a solid cylinder of radius 3 m and height 5 m oriented along the z-axis with its base at the origin is:
 A $108 \pi$ B $54 \pi$ C $90 \pi$ D $180 \pi$
GATE EE 2020   Electromagnetic Theory
Question 4 Explanation:
\begin{aligned} & \psi / \text{Crossing closed surface} \\ \oint \oint \vec{D}.\vec{ds}&=\int \int \int (\vec{\triangledown }\cdot \vec{D}) dv \\ (\vec{V}.\vec{D}) &=\frac{1}{\rho }\frac{\partial }{\partial p}(\rho D_{\rho })+\frac{1}{\rho }\frac{\partial D_{\phi }}{\partial \phi }+\frac{\partial D_{z}}{\partial z} \\ &=\frac{1}{\rho }\frac{\partial }{\partial p}(\rho 15)+\frac{1}{\rho }\frac{\partial }{\partial \phi }(2\rho )+\frac{\partial }{\partial z}(-3\rho z) \\ &=\frac{1}{\rho }15-3\rho \\ \int \int \int (\vec{\triangledown }\cdot \vec{D})dv&=\int \int \int \left ( \frac{15}{\rho}-3\rho \right )\rho d\rho d\phi dz \\ &=\int \int \int 15d\rho d\phi dz-3\int \int \int \rho ^{2}d\rho d\phi dz \\ &=15\int_{\rho =0}^{3}d\rho \int_{\phi =0}^{2\pi }d\phi \int_{z=0}^{5}-3\int_{\rho =0}^{3}\rho^{2}d\rho \int_{\phi =0}^{2\pi }d\phi \int_{z=0}^{5}dz \\ &= 15(3-0)(2\pi )(5)-3\left ( \frac{3^{3}}{3} \right )\times (2\pi )(5) \\ &=45(10\pi )-27(10\pi )=180\pi C \end{aligned}
 Question 5
The static electric field inside a dielectric medium with relative permittivity, $\varepsilon _r = 2.25$, expressed in cylindrical coordinate system is given by the following expression

$E=a_r 2r+a_\varphi \left ( \frac{3}{r} \right )+a_z6$

where $a_r,a_\varphi ,a_z$ are unit vectors along $r, \varphi \; and \;z$ directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, $\varepsilon _0$, in SI units is given by:
 A $3\varepsilon _0$ B $4\varepsilon _0$ C $5\varepsilon _0$ D $9\varepsilon _0$
GATE EE 2020   Electromagnetic Theory
Question 5 Explanation:
$\vec{D}=\epsilon \vec{E}=\epsilon _{o}\epsilon _{r}\vec{E}$
$\vec{D}=\epsilon _{o}2.25\left ( 2r\hat{a_{r}}+\frac{3}{r}\hat{a_{\phi}+6\hat{a_{z}}} \right )$
$\vec{D}=4.5\epsilon _{o}r\hat{a_{r}}+\frac{6.75\epsilon _{o}}{r}\hat{a_{\phi }}+13.5\epsilon _{o}\hat{a_{z}}$
volume charge density
$\rho _{v}=\vec{\bigtriangledown }\cdot \vec{D}$
$\rho _{v}=\frac{1}{r}\frac{\partial }{\partial r}(rD_{r})+\frac{1}{r}\frac{\partial D_{\phi }}{\partial \phi }+\frac{\partial D_{z}}{\partial z}$
$\rho _{v}=\frac{1}{r}\frac{\partial }{\partial r}(r4.5\epsilon _{o}r)+\frac{1}{r}\frac{\partial }{\partial \phi }\left ( \frac{6.75\epsilon _{o}}{r} \right )+\frac{\partial }{\partial z}(13.5\epsilon _{o})$
$\, \, \, =\frac{1}{r}\frac{\partial }{\partial r}(4.5\epsilon _{o}r^{2})+0+0$
$\, \, \, =\frac{1}{r}(4.5\epsilon _{o})(2r)=9\epsilon _{o}$
 Question 6
A co-axial cylindrical capacitor shown in Figure (i) has dielectric with relative permittivity $\varepsilon _{r1}=2$. When one-fourth portion of the dielectric is replaced with another dielectric ofrelative permittivity $\varepsilon _{r2}$, as shown in Figure (ii), the capacitance is doubled. The value of $\varepsilon _{r12}$ is ____. A 10 B 7 C 15 D 18
GATE EE 2019   Electromagnetic Theory
Question 6 Explanation:
Co-axial cylindrical capacitor-1 \begin{aligned} C_1 &=\frac{2 \pi \in _h}{\ln\left ( \frac{b}{a} \right )}=\frac{2 \pi \in _0 (2)h}{\ln \left ( \frac{R}{r} \right )}\\ C_1&=\frac{2 \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\;\;...(i) \end{aligned}
Co-axial cylindrical capacitor-2 \begin{aligned} C_2 &=\frac{\left (\frac{3 \pi}{2} \right ) \in _0 (2)h}{\ln \left ( \frac{R}{r} \right )}+\frac{\frac{\pi}{2} \in _0 \in _{r_2}h}{\ln \left ( \frac{R}{r} \right )}\\ C_2&=\frac{ \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\left [ 3+\frac{\in _{r_2}}{2} \right ]\;\;...(ii) \end{aligned}
Given, $C_2=2C_1$ ....(iii)
Put equation (i), (ii) in equation (iii),
\begin{aligned} &\frac{ \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\left [ 3+\frac{\in _{r_2}}{2} \right ]=2\left [ \frac{4 \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )} \right ]\\ &3+\frac{\in {r_2}}{2}=8\\ &\Rightarrow \; \frac{\in {r_2}}{2}=5\; \Rightarrow \; \in {r_2}=10 \end{aligned}
 Question 7
The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF.Neglecting the fringing effects, the relative permittivity of the dielectric is _____________ (up to 2 decimal places).
 A 1.55 B 2.53 C 3.75 D 4.25
GATE EE 2018   Electromagnetic Theory
Question 7 Explanation:
Given, $C=\frac{\varepsilon _0A}{d}=60pF$
In second case: Capacitance,
\begin{aligned} C_1 &=\frac{\varepsilon _0A}{d/2} =\frac{2\varepsilon _0A}{d}\\ &= 2 \times (60 pF)=120pF\\ C_2&=\frac{2\varepsilon _0\varepsilon _rA}{d} \\ &=(2 \times 60)\varepsilon _r \;pF =120\varepsilon _r \; pF\\ C_{eq}&=\frac{C_1C_2}{C_1+C_2}=\frac{120 \times 120\varepsilon _r}{(120+120\varepsilon _r)}pF \\ &= 86\;pF \;\;(given)\\ 86&=\frac{120 \times 120\varepsilon _r}{120(1+\varepsilon _r)} \\ \frac{86}{120}&= \frac{\varepsilon _r}{1+\varepsilon _r}\\ \varepsilon _r &=\frac{86}{34}=2.53 \end{aligned}
 Question 8
A positive charge of 1 nC is placed at (0,0,0.2) where all dimensions are in metres. Consider the x-y plane to be a conducting ground plane. Take $\epsilon _{0}=8.85 \times 10^{-12}$ F/m. The z component of the E field at (0,0,0.1) is closest to
 A 899.18 V/m B $-899.18 V/m$ C 999.09 V/m D $-999.09 V/m$
GATE EE 2018   Electromagnetic Theory
Question 8 Explanation: Net electric field at point P due to charge Q is,
\begin{aligned} \vec{E_{12}}&=\frac{Q\vec{R_{12}}}{4 \pi \varepsilon _0|\vec{R_{12}}|^3} \\ &= \frac{1 \times 10^{-9}(-0.1\hat{a_z})}{4 \pi (8.854 \times 10^{-12})(0.1)^3}\\ &+ \frac{-1 \times 10^{-9}(-0.3\hat{a_z})}{4 \pi (8.854 \times 10^{-12})(0.3)^3}\\ &= \left [ \frac{-10^5}{4 \pi (8.854)}- \frac{10^5}{4 \pi (8.854)9} \right ]\hat{a_z}\\ &=(-898.774-99.863) \hat{a_z}\\ &= -999.09\hat{a_z}\; V/m \end{aligned}
 Question 9
A thin soap bubble of radius R = 1 cm, and thickness $a=3.3\mu m(a\lt \lt R)$, is at a potential of 1 V with respect to a reference point at infinity. The bubble bursts and becomes a single spherical drop of soap (assuming all the soap is contained in the drop) of radius r. The volume of the soap in the thin bubble is $4\pi R^{2}a$ and that of the drop is $\frac{4}{3} \pi r^{3}$. The potential in volts, of the resulting single spherical drop with respect to the same reference point at infinity is __________. (Give the answer up to two decimal places.) A 10.03 B 20.25 C 30.28 D 40.08
GATE EE 2017-SET-2   Electromagnetic Theory
Question 9 Explanation:
After burst,
$4 \pi R^2 a=\frac{4}{3}\pi r^3$
Radius of soap drop $=(3 R^2a)^{1/3}=0.099665cm$
Initial voltage was 1 V and $C=4 \pi \varepsilon _0R$
and initial charge, $Q=(4 \pi \varepsilon _0R \times 1)$
Since after bursting Q remainsame and $C=4 \pi \varepsilon _0r$
New potential on soap drop,
$V=\frac{Q}{C}=\frac{4 \pi \varepsilon _0R}{4 \pi \varepsilon _0r}=\frac{1}{0.099665}=10.03V$
 Question 10
Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million electrons are added to this sphere, these electrons will be distributed.
 A uniformly over the entire volume of the sphere B uniformly over the outer surface of the sphere C concentrated around the centre of the sphere D along a straight line passing through the centre of the sphere
GATE EE 2017-SET-2   Electromagnetic Theory
Question 10 Explanation:
Added charge (one million electrons) to be solid spherical conductor is uniformly distributed over the outer surface of the sphere.
There are 10 questions to complete.