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

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 $\mu C$

In the circuit shown in figure, energy stored in \(6~\mu\text{F}\) 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) $2eV/m$

The dimension of (1/2) ${\epsilon }_0{E}^2 ({\epsilon }_0$: permittivity of free space; E: electric field) is

(1) MLT^–1

(2) ML²L^–2

(3) ML^–1T^–2

(4) ML²T^–1

Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at P is given by (r >> 2a) (Where p = 2qa) 

(1) $V=\frac{p\cos \theta }{4\pi {\epsilon }_0{r}^2}$

(2) $V=\frac{p\cos \theta }{4\pi {\epsilon }_{0}r}$

(3) $V=\frac{p\sin \theta }{4\pi {\epsilon }_{0}r}$

(4) $V=\frac{p\cos \theta }{2\pi {\epsilon }_0{r}^2}$ 

Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field E pointing co-parallel to the positive direction of the x-axis. The coordinates of the points P, Q, R, and S are $(a,b,0),(2a,0,0),(a,-b,0)$ and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression 

(1) qEa

(2) – qEa

(3) $qEa\sqrt{2}$

(4) $qE\sqrt{[{\left(2a\right)}^2+b^2]}$