The variation of potential with distance R from a fixed point is as shown below. The electric field at R = 5 m is 

(1) 2.5 volt/m

(2) –2.5 volt/m

(3) 2/5 volt/m

(4) –2/5 volt/m

A charge +q is fixed at each of the points $\mathrm{x}={\mathrm{x}}_0, \mathrm{x}=3{\mathrm{x}}_0, \mathrm{x}=5{\mathrm{x}}_0$, ........... upto $\infty$ on X-axis and charge –q is fixed on each of the points $\mathrm{x}=2{\mathrm{x}}_0, \mathrm{x}=4{\mathrm{x}}_0, \mathrm{x}=6{\mathrm{x}}_0$,............. upto $\infty$. Here ${\mathrm{x}}_0$ is a positive constant. Take the potential at a point due to a charge Q at a distance r from it to be $\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}$. Then the potential at the origin due to above system of charges will be

1. zero                           

2. $\frac{\mathrm{q}}{8{\mathrm{\pi \epsilon }}_0{\mathrm{x}}_0 {\log }_{\mathrm{e}} 2}$

3. $\infty$                             

4. $\frac{\mathrm{q} {\log }_{\mathrm{e}} 2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{x}}_0}$

A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is 

(1) Zero

(2) $\frac{25C{V}^2}{6}$

(3) $\frac{3C{V}^2}{2}$

(4) $\frac{9C{V}^2}{2}$

The electric potential at a point (x, y) in the x – y plane is given by V = –kxy. The field intensity at a distance r from the origin varies as 

(1) r²

(2) r

(3) $\frac{1}{r}$

(4) $\frac{1}{r^2}$

Four identical particles each of mass m and charge q are kept at the four corners of a square of length L. The final velocity of these particles after setting them free will be.

1. ${\left[\frac{{\mathrm{Kq}}^2}{\mathrm{mL}}\left(5.4\right)\right]}^{1/2}$                           

2. ${\left[\frac{{\mathrm{Kq}}^2}{\mathrm{mL}}\left(1.35\right)\right]}^{1/2}$

3. ${\left[\frac{{\mathrm{Kq}}^2}{\mathrm{mL}}\left(2.7\right)\right]}^{1/2}$                           

4. Zero

Three uncharged capacitors of capacities ${\mathrm{C}}_1, {\mathrm{C}}_2$ and ${\mathrm{C}}_3$ are connected to one another as shown in the figure.

Points A, B, and D are at potential ${\mathrm{V}}_1, {\mathrm{V}}_2$ and ${\mathrm{V}}_3$ ,then the potential at O will be 

1. $\frac{{\mathrm{V}}_1{\mathrm{C}}_1+{\mathrm{V}}_2{\mathrm{C}}_2+{\mathrm{V}}_3{\mathrm{C}}_3}{{\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3}$                   

2. $\frac{{\mathrm{V}}_1+{\mathrm{V}}_2+{\mathrm{V}}_3}{{\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3}$

3. $\frac{{\mathrm{V}}_1\left({\mathrm{V}}_2+{\mathrm{V}}_3\right)}{{\mathrm{C}}_1\left({\mathrm{C}}_2+{\mathrm{C}}_3\right)}$                               

4. $\frac{{\mathrm{V}}_1{\mathrm{V}}_2{\mathrm{V}}_3}{{\mathrm{C}}_1{\mathrm{C}}_2{\mathrm{C}}_3}$

Figure shows a solid hemisphere with a charge of 5nC distributed uniformly through its volume. The hemisphere lies on a plane and point P is located on the plane, along a radial line from the centre of curvature at distance 15 cm. The electric potential at point P due to the hemisphere, is –

1.   150 V                                       

2.   300 V

3.   450 V                                       

4.   600 V

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. The magnitude and direction of the electric field are given by: