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. $\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right) \mathrm{V}/\mathrm{m}$                           

2. $\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}\right) \mathrm{V}/\mathrm{m}$

3. $\left(-\hat{\mathrm{i}}+\hat{\mathrm{j}}\right) \mathrm{V}/\mathrm{m}$                         

4. $\left(-\hat{\mathrm{i}}-\hat{\mathrm{j}}\right) \mathrm{V}/\mathrm{m}$

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. $\mathrm{V}-\mathrm{E}2\mathrm{R} \sin \frac{\mathrm{\theta }}{2}$

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

4. $\mathrm{V}+2\mathrm{ER} \sin \frac{\mathrm{\theta }}{2}$

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.  $\frac{400}{3} \mathrm{\mu C}$                               

2.  $100 \mathrm{\mu C}$

3.  $\frac{100}{3} \mathrm{\mu C}$                               

4.  $50 \mathrm{\mu C}$

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.  $3.2\times {10}^{-17} \mathrm{J}$                      

2.  $-3.2\times {10}^{-17} \mathrm{J}$

3.  $200 \mathrm{J}$                                     

4.  $-200 \mathrm{J}$

Two charged conducting spheres of radii ${\mathrm{r}}_1$ and ${\mathrm{r}}_2$ are at the same potential. The ratio of their surface charge densities will be -

1.  ${\mathrm{r}}_1/{\mathrm{r}}_2$                               

2.  ${\mathrm{r}}_2/{\mathrm{r}}_1$

3.  ${{\mathrm{r}}_1}^2/{{\mathrm{r}}_2}^2$                           

4.  ${{\mathrm{r}}_2}^2/{{\mathrm{r}}_1}^2$

At the mid point of a line joining an electron and a proton, the values of E and V will be.

1. $\mathrm{E}=0, \mathrm{V}\ne 0$                   

2. $\mathrm{E}\ne 0, \mathrm{V}=0$

3. $\mathrm{E}\ne 0, \mathrm{V}\ne 0$                   

4. $\mathrm{E}=0, \mathrm{V}=0$

A charge of $10\mu \mathrm{C}$ is kept at the origin of $\mathrm{X}–\mathrm{Y}$ coordinate system. The potential difference in volts between two points (a, 0) and $\left(\mathrm{a}/\sqrt{2} , \mathrm{a}/\sqrt{2}\right)$ will be.

1.  Zero                             

2.  $9\times {10}^4$

3.  $\frac{9\times {10}^4}{\mathrm{a}}$                           

4.  $\frac{9\times {10}^4}{\sqrt{2}}$

Find equivalent capacitance between $\mathrm{X}$ and $\mathrm{Y}$ if each capacitor is $4 \mathrm{\mu F}$.

1.  $4 \mathrm{\mu F}$                                     

2.  $2 \mathrm{\mu F}$

3.  $12 \mathrm{\mu F}$                                   

4.  $1 \mathrm{\mu F}$