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
V(r)=\frac{cd(V_2-V_1)}{(d-c)r}-\frac{V_1c+V_2d-2V_1d}{d-c} | |
V(r)=\frac{cd(V_1-V_2)}{(d-c)r}-\frac{V_2d-V_1c}{d-c} | |
V(r)=\frac{cd(V_1-V_2)}{(d-c)r}-\frac{V_1c-V_2c}{d-c} | |
V(r)=\frac{cd(V_2-V_1)}{(d-c)r}-\frac{V_2c-V_1c}{d-c} |
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)}
\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 ________.
2 | |
5 | |
6 | |
8 |
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.
Take \dfrac{1}{4\pi \varepsilon _{0}}=9\times 10^{9} in SI units. All coordinates are in meters.
5.25 | |
9 | |
8.62 | |
2.47 |
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:
108 \pi | |
54 \pi | |
90 \pi | |
180 \pi |
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:
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:
3\varepsilon _0 | |
4\varepsilon _0 | |
5\varepsilon _0 | |
9\varepsilon _0 |
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}
\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}
There are 5 questions to complete.