A hollow conducting sphere is placed in an electric field produced by a point charge placed at \(P\) as shown in the figure. Let\(​​V_A ~,V_B~,V_C\) be the potentials at points \(A\), \(B\) and \(C\) respectively. Then:

        
1. \(V_A<V_B<V_C\)
2. \(V_A>V_B>V_C\)
3. \(V_C>V_B=V_A\)
4. \(V_A=V_B=V_C\)

Four particles each having charge q are placed at the vertices of a square of side a. The value of the electric potential at the midpoint of one of the side will be

1.  0

2.  $\frac{1}{4\pi {ϵ}_0}\frac{2q}{a}(2+\frac{2}{\sqrt{5}})$

3.  $\frac{1}{4\pi {ϵ}_0}\frac{2q}{a}(2-\frac{2}{\sqrt{5}})$

4.  $\frac{1}{4\pi {ϵ}_0}\frac{2q}{a}(1+\frac{1}{\sqrt{5}})$

If E be the electric field inside a parallel plate capacitor due to Q and -Q charges on the two plates, then electrostatic force on plate having charge -Q due to the plate having charge +Q will be

(1) -QE

(2) $\frac{-QE}{2}$

(3) QE

(4) $\frac{-QE}{4}$

A metallic sphere of capacitance $C_1$, charged to electric potential $V_1$ is connected by a metal wire to another metallic sphere of capacitance $C_2$ charged to electric potential $V_2$. The amount of heat produced in connecting the wire during the process is:

1.$\frac{C_1{C}_2}{2\left(C_1+C_2\right)}{\left(V_1+V_2\right)}^2$

2. $\frac{C_1{C}_2}{2(C_1+C_2)}{(V_1-V_2)}^2$

3. $\frac{C_1{C}_2}{C_1+C_2}{(V_1-V_2)}^2$

4. zero

The electric potential at the surface of a charged solid sphere of insulator is 20V. The value of electric potential at its centre will be

1.  30V

2.  20V

3.  40V

4.  Zero

The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then new capacitance become

1. $\frac{KC}{2\left(K+1\right)}$

2. $\frac{2KC}{K+1}$

3. $\frac{5KC}{4K+1}$

4. $\frac{4KC}{3K+1}$

The electric potential at a point at distance 'r' from a short dipole is proportional to

(1) $r^2$

(2) ${\mathrm{r}}^{-1}$

(3) $r^{-2}$

(4) $r^1$

A hollow charged metal spherical shell has radius R. If the potential difference between its surface and a point at a distance 3R from the center is V, then the value of electric field intensity at a point at distance 4R from the center is

1.  $\frac{3\mathrm{V}}{19\mathrm{R}}$

2.  $\frac{\mathrm{V}}{6\mathrm{R}}$

3.  $\frac{3\mathrm{V}}{32\mathrm{R}}$

4.  $\frac{3\mathrm{V}}{16\mathrm{R}}$

Capacitors $C_1=10\mu F and C_2=30\mu F$ are connected in series across a source of emf 20KV. The potential difference across $C_1$ will be

(1) 5 KV

(2) 15 KV

(3) 10 KV

(4) 20 KV