Two identical metal plates are given charges $Q_1 and Q_2\left(<Q_1\right)$ respectively. If they are now brought close to form a parallel plate capacitor with capacitance $C$, the potential difference between them is :

1. $\frac{Q_1+Q_2}{2C}$

2. $\frac{Q_1+Q_2}{C}$

3. $\frac{Q_1-Q_2}{C}$

4. $\frac{Q_1-Q_2}{2C}$

A and B are two concentric metallic shells. If A is positively charged and B is earthed, then electric

1.  Field at common centre is non-zero

2.  Field outside B is nonzero

3.  Potential outside B is positive

4.  Potential at common centre is positive

$\mathrm{The} \mathrm{electric} \mathrm{potential} \mathrm{at} \mathrm{a} \mathrm{point} \mathrm{at} \mathrm{distance} \sqrt{3}\mathrm{R} \mathrm{from} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{disc}$
$\mathrm{of} \mathrm{radius} \mathrm{R} \mathrm{lying} \mathrm{in} \mathrm{the} \mathrm{axis} \mathrm{of} \mathrm{the} \mathrm{disc} \mathrm{whose} \mathrm{surface} \mathrm{charge} \mathrm{density} \mathrm{is} \mathrm{\sigma }$
$\mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}:$
$1. \frac{\sigma }{2{\epsilon }_0}\left[2-\sqrt{3}\right]R 2. \frac{\sigma }{2{\epsilon }_0}\left[2+\sqrt{3}\right]R$
$3.\frac{\sigma }{2{\epsilon }_0}\left[\sqrt{3}-\sqrt{2}\right]R 4. \frac{\sigma }{2{\epsilon }_0}\left[\sqrt{3}+\sqrt{2}\right]R$

A 4 $\mathrm{\mu F}$ capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 $\mathrm{\mu F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

1. $2.67\times {10}^{-2} \mathrm{J}$

2. $4.99\times {10}^{-2} \mathrm{J}$

3. $6.67\times {10}^{-2} \mathrm{J}$

4. $8.84\times {10}^{-2} \mathrm{J}$

An air capacitor of capacity $C=10\mu F$ is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from battery to the capacitor is

1. 120 $\mu C$

2. 699 $\mu C$

3. 480 $\mu C$

4. 24 $\mu C$

The figure shows some of the equipotential surfaces. Magnitude and direction of the electric field is given by

1.  200 V/m, making an angle ${120}^0$ with the x-axis

2. 100 V/m, pointing towards the negative x-axis

3. 200 V/m, making an angle $-{60}^0$ with the x-axis

4. 100 V/m, making an angle ${30}^0$ with the x-axis

In the circuit shown in figure, energy stored in 6 $\mathrm{\mu F}$ capacitor will be:
 

Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work done in removing a charge – Q from its centre to infinity is 

(1) 0

(2) $\frac{\sqrt{2}{Q}^2}{4\pi {\epsilon }_{0}a}$

(3) $\frac{\sqrt{2}{Q}^2}{\pi {\epsilon }_{0}a}$

(4) $\frac{Q^2}{2\pi {\epsilon }_{0}a}$

Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of the electric field at the surface of the spheres is 

(1) a/b

(2) b/a

(3) a²/b²

(4) b²/a²

An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed of the electron will be 

(1) $V\sqrt{e/m}$

(2) $\sqrt{eV/m}$

(3) $\sqrt{2eV/m}$

(4) $2eV/m$