Question 1 |

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 1 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 2 |

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 2 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}

Question 3 |

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 ____.

10 | |

7 | |

15 | |

18 |

Question 3 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}

\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 4 |

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).

1.55 | |

2.53 | |

3.75 | |

4.25 |

Question 4 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}

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 5 |

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

899.18 V/m | |

-899.18 V/m | |

999.09 V/m | |

-999.09 V/m |

Question 5 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 6 |

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.)

10.03 | |

20.25 | |

30.28 | |

40.08 |

Question 6 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

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 7 |

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.

uniformly over the entire volume of the sphere | |

uniformly over the outer surface of the sphere | |

concentrated around the centre of the sphere | |

along a straight line passing through the centre of the sphere |

Question 7 Explanation:

Added charge (one million electrons) to be solid spherical conductor is uniformly distributed over the outer surface of the sphere.

Question 8 |

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?

The particles will never collide | |

All will collide together | |

Proton and Neutron | |

Electron and Neutron |

Question 8 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.

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.

Question 9 |

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

A | |

B | |

C | |

D |

Question 9 Explanation:

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

Question 10 |

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

1 | |

2 | |

4 | |

16 |

Question 10 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.

E=\frac{V}{d}= constant for both regions.

There are 10 questions to complete.