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

🚀 Solve in Practice Mode

📖 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 ________.

🚀 Solve in Practice Mode

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

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.

🚀 Solve in Practice Mode

📖 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}

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:

🚀 Solve in Practice Mode

📖 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:

🚀 Solve in Practice Mode

📖 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 ____.