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

A parallel plate capacitor with air between the plates is charged to a potential difference of 500V and then insulated. A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes 75V. The dielectric constant of plastic is –

1.  10/3                           

2.  5 

3.  20/3                           

4.  10

The circuit was in the shown state for a long time. Now if the switch S is closed then the charge that flows through the switch S, will be –

1.  $\frac{400}{3}<mspace\; width="1em"></mspace>\mathrm{\mu C}$                               

2.  $100<mspace\; width="1em"></mspace>\mathrm{\mu C}$

3.  $\frac{100}{3}<mspace\; width="1em"></mspace>\mathrm{\mu C}$                               

4.  $50<mspace\; width="1em"></mspace>\mathrm{\mu C}$

The potential at a certain point in an electric field is 200 V. The work done in carrying an electron upto that point will be.

1.  $3.2\times {10}^{-17}<mspace\; width="1em"></mspace>\mathrm{J}$                      

2.  $-3.2\times {10}^{-17}<mspace\; width="1em"></mspace>\mathrm{J}$

3.  $200<mspace\; width="1em"></mspace>\mathrm{J}$                                     

4.  $-200<mspace\; width="1em"></mspace>\mathrm{J}$