GATE EE · Electromagnetic Theory

Electrostatic Fields – GATE EE Electromagnetic TheoryPractice Questions & PYQs

Generate GATE-level questions on Electrostatics. Focus on: 1. Coulomb's Law and Electric Field Intensity. 2. Gauss's Law and its applications. 3. Electric Potential, Capacitance, and Energy in electrostatic fields.

35 questions · 20 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2025MCQ

An air filled cylindrical capacitor (capacitance $\mathrm{C}_{0}$ ) of length $L$ , with $a$ and $b$ as its inner and outer radii, respectively, consists of two coaxial conducting surfaces. Its cross-sectional view is shown in Fig. (i). In order to increase the capacitance, a dielectric material of relative permittivity $\varepsilon_{\mathrm{r}}$ is inserted inside $50 \%$ of the annular region as shown in Fig. (ii). The value of $\varepsilon_{\mathrm{r}}$ for which the capacitance of the capacitor in Fig. (ii), becomes $5 \mathrm{C}_{0}$ is

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📖 Explanation

Step 1: Capacitance of a cylindrical capacitor The capacitance of a coaxial cylindrical capacitor filled with dielectric of permittivity $\varepsilon$ is given by: $C = \frac{2\pi \varepsilon L}{\ln(b/a)}$ So, for Fig. (i) with air throughout (i.e., permittivity $\varepsilon_0$ ): $C_0 = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$ Step 2: For Fig. (ii) The cross-section is split into two equal parts (in parallel): One half is filled with air $\varepsilon_0$ The other half is filled with dielectric $\varepsilon_r \varepsilon_0$ Since they are in parallel, total capacitance is: $C = \frac{1}{2} C_0 + \frac{1}{2} \varepsilon_r C_0 = \frac{C_0}{2}(1 + \varepsilon_r)$ We are given that $C = 5C_0$ . So: $\frac{C_0}{2}(1 + \varepsilon_r) = 5C_0$ Divide both sides by $C_0$ : $\frac{1 + \varepsilon_r}{2} = 5$ $1 + \varepsilon_r = 10 \Rightarrow \varepsilon_r = 9$ Final Answer: (C) 9

Q2GATE 2025MCQ

Which one of the following figures represents the radial electric field distribution ${E}_{{R}}$ caused by a spherical cloud of electrons with a volume charge density, $\rho=-3 \rho_{0}$ for $0 \leq R \leq a$ (both $\rho_{0}$ , a are positive and $R$ is the radial distance) and $\rho=0$ for ${R} > {a}$ ?

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📖 Explanation

$0 < \mathrm{R} < \mathrm{a}:$ $\oiint \overline{\mathrm{E}} \cdot \overline{\mathrm{dA}}=\frac{\mathrm{Q}_{\mathrm{ENC}}}{\varepsilon_{0}}=\frac{1}{\varepsilon_{0}} \iiint\left(-3 \rho_{0}\right) \mathrm{dv}$ $\mathrm{E}_{\mathrm{r}} 4 \pi \mathrm{R}^{2}=\frac{1}{\varepsilon_{0}}\left(-3 \rho_{0}\right) \frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{E}_{\mathrm{r}}=\frac{-3 \rho_{0}}{\varepsilon_{0}} \frac{\mathrm{R}}{3}=\frac{-\rho_{0} \mathrm{R}}{\varepsilon_{0}}$

Explanation Figure 2

Q3GATE 2022MCQ

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

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📖 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), $V=-\frac{(V_2-V_1)cd}{r(d-c)}+V_1+\frac{(V_2-V_1)cd}{(d-c)}$ $V=\frac{(V_1-V_2)cd}{r(d-c)}+\frac{(V_2d-V_1c)}{(d-c)}$

Q4GATE 2021NAT

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 ________.

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📖 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), $\Rightarrow \Rightarrow \qquad \qquad \frac{E_{2}}{E_{1}}=5$

Explanation Figure 1

Q5GATE 2021NAT

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.

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📖 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}

Explanation Figure 1

Q6GATE 2020MCQ

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:

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📖 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}

Q7GATE 2020MCQ

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:

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📖 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 $\rho _{v}=\vec{\bigtriangledown }\cdot \vec{D}$ $\rho _{v}=\frac{1}{r}\frac{\partial }{\partial r}(rD_{r})+\frac{1}{r}\frac{\partial D_{\phi }}{\partial \phi }+\frac{\partial D_{z}}{\partial z}$ $\rho _{v}=\frac{1}{r}\frac{\partial }{\partial r}(r4.5\epsilon _{o}r)+\frac{1}{r}\frac{\partial }{\partial \phi }\left ( \frac{6.75\epsilon _{o}}{r} \right )+\frac{\partial }{\partial z}(13.5\epsilon _{o})$ $\, \, \, =\frac{1}{r}\frac{\partial }{\partial r}(4.5\epsilon _{o}r^{2})+0+0$ $\, \, \, =\frac{1}{r}(4.5\epsilon _{o})(2r)=9\epsilon _{o}$

Q8GATE 2019NAT

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 ____.

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📖 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}

Explanation Figure 2

Explanation Figure 3

Q9GATE 2018NAT

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).

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📖 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}

Explanation Figure 1

Q10GATE 2018MCQ

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

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📖 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}

Explanation Figure 1

Q11GATE 2017MCQ

Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the center of the line joining the electron and proton. At t=0, the particles are released but are constrained to move along the same straight line. Which of these will collide first?

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📖 Explanation

Given that electron, neutron and proton are in straight line. The electron will move towards proton and proton will move towards electron and force will be same $F=\frac{q_1q_2}{4 \pi \in _0 R^2}$ . But acceleration of electron will be more than proton as mass of electron $\lt$ mass of proton. Since neutron are neutral they will not move. Thus electron will hit neutron first.

Explanation Figure 1

Q12GATE 2017NAT

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.)

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📖 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, $V=\frac{Q}{C}=\frac{4 \pi \varepsilon _0R}{4 \pi \varepsilon _0r}=\frac{1}{0.099665}=10.03V$

Q13GATE 2017MCQ

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.

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Q14GATE 2016MCQ

Two electric charges q and -2q are placed at (0,0) and (6,0) on the x-y plane. The equation of the zero equipotential curve in the x-y plane is

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📖 Explanation

Charge, Q is located at (0,0) and $-2\theta$ is located at (6,0). To find V at any point $(x,y)$

\begin{aligned} V_Q &= \frac{Q}{4 \pi \in \sqrt{x^{2+}y^2}}\\ V_{-2Q}&=\frac{-2Q}{4 \pi \in (\sqrt{(x-6)^2 +y^2})} \\ V_{total}&=0 \\ &= \frac{Q}{4 \pi \in \sqrt{x^{2+}y^2}} \\ & +\frac{-2Q}{4 \pi \in (\sqrt{(x-6)^2 +y^2})} \\ \sqrt{(x-6)^2 +y^2}&=2(\sqrt{x^{2+}y^2}) \\ x^{2+}36-12x+y^2&=4x^{2+}4y^2 \\ 3x^{2+}3y^{2+}12x&=36 \\ x^{2+}y^{2+}4x&=12 \\ (x+2)^{2+}y^2&=16 \end{aligned}

Q15GATE 2016MCQ

Two electrodes, whose cross-sectional view is shown in the figure below, are at the same potential. The maximum electric field will be at the point

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📖 Explanation

At the point C, $\bar{E}$ (electric field intensity) is maximum being closest to the other plate.

Explanation Figure 2

Q16GATE 2016MCQ

A parallel plate capacitor filled with two dielectrics is shown in the figure below. If the electric field in the region A is 4 kV/cm, the electric field in the region B, in kV/cm, is

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📖 Explanation

As voltage is same across both the regions and distance between two plates is also same, then electric field remains same throughout the both region. As we know, $E=\frac{V}{d}=$ constant for both regions.

Q17GATE 2015MCQ

Two semi-infinite dielectric regions are separated by a plane boundary at y = 0. The dielectric constants of region 1 (y $\lt$ 0) and region 2 (y $\gt$ 0) are 2 and 5, respectively. Region 1 has uniform electric field $\vec{E}=3\hat{a}_{x}+4\hat{a}_{y}+2\hat{a}_{z}$ , where $\hat{a}_{x},\hat{a}_{y}$ and $\hat{a}_{z}$ are unit vectors along the x, y and z axes, respectively. The electric field in region 2 is

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📖 Explanation

Given that, at the interface (y=0) there is no surface charge. Region 1 ( $y \lt 0$ ) has $\varepsilon _r=2$ Region 2 ( $y \gt 0$ ) has $\varepsilon _r=5$ Electric field in region 1 is $3a_x+4a_y+2a_z$ Normal component of electric field is $4a_y$ tangential component of electric field is $3a_x+2a_y$ Now, $E_1=E_{1t}+E_{1n}$ The tangentail component of $E_1=$ tangentail component of $E_2$ $E_{1t}=E_{2t}=3a_x+2a_z$

\begin{aligned} \varepsilon _1 E_{1n}&=\varepsilon _2 E_{2n}\\ E_{2n}&=\frac{\varepsilon _1 \times 4a_y}{\varepsilon _2} \\ &= \frac{2 \times 4a_y}{5}=1.6a_y\\ E_2&=E_{2t}+E_{2n}\\ &=3a_x+1.6a_y+2a_z \end{aligned}

Q18GATE 2015NAT

A parallel plate capacitor is partially filled with glass of dielectric constant 4.0 as shown below. The dielectric strengths of air and glass are 30 kV/cm and 300 kV/cm, respectively. The maximum voltage (in kilovolts), which can be applied across the capacitor without any breakdown, is ______.

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📖 Explanation

A capacitor is made as shown in figure Breakdown of dielectric air is 30 kV/cm and breakdown of dielectric glass is 300 kV/cm. We need to find maximum value of $V_{AB}$ so that no $C_1$ or $C_2$ breaks down. Since, area and width of $C_1$ and $C_2$ is same but $\varepsilon _{r_1}=1$ and $\varepsilon _{r_2}=4$ So, $C_2=4C_1$ So, if $V_{AB}$ is applied, $V_{C_1}=\frac{4}{5}V_{AB}$ and $V_{C_2}=\frac{1}{5}V_{AB}$ For $C_1$ not breakdown, we need

\begin{aligned} \frac{4}{5}\frac{V_{AB}}{5mm}&\lt 30kV/cm \\ V_{AB}& \lt \frac{30 \times 25 \times 1000}{4}V \\ V_{A}& \lt 18.75kV \end{aligned}

Explanation Figure 2

Q19GATE 2015MCQ

Two semi-infinite conducting sheets are placed at right angles to each other as shown in the figure. A point charge of +Q is placed at a distance of d from both sheets. The net force on the charge is $\frac{Q^{2}}{4\pi \epsilon _{0}}\frac{K}{d^{2}}$ , where K is given by

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📖 Explanation

Due to image of the charge, three new charges will be created as shown in figure Total force (4) due to (1), (2), (3) will be

\begin{aligned} F_{total}&=F_{41}+F_{42}+F_{43}\\ F_{41}&=\text{Force as (4) due to (1)}\\ &=-\frac{Q^2}{4 \pi \varepsilon _0}\frac{1}{(4d^2)}\hat{j}\\ F_{42}&=\text{Force as (4) due to (2)}\\ &=+\frac{Q^2}{4 \pi \varepsilon _0}\frac{1}{(8d^2)}\left ( \frac{\hat{j}}{\sqrt{2}} + \frac{\hat{i}}{\sqrt{2}} \right )\\ F_{43}&=\text{Force as (4) due to (3)}\\ &=-\frac{Q^2}{4 \pi \varepsilon _0}\frac{1}{(4d^2)}\hat{i}\\ F_{total}&=\frac{Q^2}{4 \pi \varepsilon _0}\left ( \frac{1-2\sqrt{2}}{8\sqrt{2}}\hat{i}+ \frac{1-2\sqrt{2}}{8\sqrt{2}}\hat{j} \right ) \end{aligned}

Explanation Figure 2

Q20GATE 2014MCQ

A hollow metallic sphere of radius r is kept at potential of 1 Volt. The total electric flux coming out of the concentric spherical surface of radius R( $\gt$ r) is

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