A uniform electric field of 400 V/m exists in space as shown in graph. Two points A and B are also shown with their co-ordinates. The potential difference ${\mathrm{V}}_{\mathrm{B}}<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>{\mathrm{V}}_{\mathrm{A}}$ in volts, is – 

1.   18 V                           

2.   15 V

3.   8 V                             

4.   12 V

Figure shows three circular arcs, each of radius R and total charge as indicated. The net electric potential at the centre of curvature is –

1. $\frac{\mathrm{Q}}{2{\mathrm{\pi \epsilon }}_0\mathrm{R}}$                               

2. $\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}}$

3. $\frac{2\mathrm{Q}}{{\mathrm{\pi \epsilon }}_0\mathrm{R}}$                                 

4. $\frac{\mathrm{Q}}{{\mathrm{\pi \epsilon }}_0\mathrm{R}}$

A neutral conducting spherical shell is kept near a charge q as shown. The potential at point P due to the induced charges is –

1. $\frac{\mathrm{kq}}{\mathrm{r}}$                                                 

2. $\frac{\mathrm{kq}}{\mathrm{r}\text{'}}$

3. $\frac{\mathrm{kq}}{\mathrm{r}}-\frac{\mathrm{kq}}{\mathrm{r}\text{'}}$                                         

4. $\frac{\mathrm{kq}}{\mathrm{CP}}$

Two concentric uniformly charged spheres of radius 10 cm and 20 cm. are arranged as shown in figure. Potential difference between the sphere is –

1.  $4.5\times {10}^{11}<mspace\; width="1em"></mspace>\mathrm{V}$                         

2.  $2.7\times {10}^{11}<mspace\; width="1em"></mspace>\mathrm{V}$

3.  $0$                                             

4.  None of these

In a uniform field – 

1. all points are at the same potential 

2. pairs of points separated by the same distance must have the same potential difference 

3. no two points can have the same potential 

4. none of the above

Two metallic bodies separated by a distance of 20 cm, are given equal and opposite charges of the magnitude of 0.88$\mu$C. The component of the electric field along the line AB, between the plates, varies as, ${\mathrm{E}}_{\mathrm{x}}=(3{\mathrm{x}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>0.4)<mspace\; width="1em"></mspace>\mathrm{N}/\mathrm{C}$ where x (in meters) is the distance from one body towards the other body as shown.

1. The capacitance of the system is 10F 

2. The capacitance of the system is 20F 

3. The potential difference between A and C is 0.088 volt 

4. The potential difference between A and C is cannot be determined from the given data 

Electrical potential ‘v’ in space as a function of coordinates is given by, $\mathrm{v}=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}$ . Then the electric field intensity at (1, 1, 1) is given by –

1.  $-\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)$                         

2.  $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

3.  zero                                           

4.  $\frac{1}{\sqrt{3}}\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)$

Two concentric, thin metallic spheres of radii ${\mathrm{R}}_1$ and ${\mathrm{R}}_2$ $\left({\mathrm{R}}_1<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>{\mathrm{R}}_2\right)$ bear changes ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ respectively. Then the potential at distance r between ${\mathrm{R}}_1$ and ${\mathrm{R}}_2$ will be $\left(\mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)$

1.  $\mathrm{k}<mspace\; width="1em"></mspace>\left(\frac{{\mathrm{Q}}_1+{\mathrm{Q}}_2}{\mathrm{r}}\right)$                       

2.  $\mathrm{k}<mspace\; width="1em"></mspace>\left(\frac{{\mathrm{Q}}_1}{\mathrm{r}}+\frac{{\mathrm{Q}}_2}{{\mathrm{R}}_2}\right)$

3.  $\mathrm{k}<mspace\; width="1em"></mspace>\left(\frac{{\mathrm{Q}}_2}{\mathrm{r}}+\frac{{\mathrm{Q}}_1}{{\mathrm{R}}_1}\right)$                     

4.  $\mathrm{k}<mspace\; width="1em"></mspace>\left(\frac{{\mathrm{Q}}_1}{{\mathrm{R}}_1}+\frac{{\mathrm{Q}}_1}{{\mathrm{R}}_2}\right)<mspace\; width="1em"></mspace>$

The grid (each square of 1m × 1m), represents a region in space containing a uniform electric field.

If potentials at points O, A, B, C, D, E, F and G, H are respectively 0, –1, –2, 1, 2, 0, –1, 1 and 0 volts, find the electric field intensity –

1. $\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)<mspace\; width="1em"></mspace>\mathrm{V}/\mathrm{m}$                           

2. $\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}\right)<mspace\; width="1em"></mspace>\mathrm{V}/\mathrm{m}$

3. $\left(-\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)<mspace\; width="1em"></mspace>\mathrm{V}/\mathrm{m}$                         

4. $\left(-\hat{\mathrm{i}}-\hat{\mathrm{j}}\right)<mspace\; width="1em"></mspace>\mathrm{V}/\mathrm{m}$

Figure shows an electric line of force which curves along a circular arc.

The magnitude of electric field intensity is same at all points on this curve and is equal to E. If the potential at A is V, then the potential at B is –

1. $\mathrm{V}-\mathrm{ER\theta }$                             

2. $\mathrm{V}-\mathrm{E}2\mathrm{R}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\frac{\mathrm{\theta }}{2}$

3. $\mathrm{V}+\mathrm{ER\theta }$                             

4. $\mathrm{V}+2\mathrm{ER}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\frac{\mathrm{\theta }}{2}$