Question

Share

Surface charge density is defined as charge per unit surface area of surface charge distribution. i.e., $\sigma=\frac{\text{dq}}{\text{dS}}.$ Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs having magnitude of 17.0 × 10^-22Cm^-2 as shown. The intensity of electric field at a point is $\text{E}=\frac{\sigma}{\in_0},$ where$\in_0=$ permittivity of free space.

1. E in the outer region of the first plate is:

1. 17 × 10^-22 N/C
2. 1.5 × 10^-25 N/C
3. 1.9 × 10^-10 N/C
4. Zero.

2. E in the outer region of the second plate is:

1. 17 × 10^-22 N/C
2. 1.5 × 10^-15 N/C
3. 1.9 × 10^-10 N/C
4. Zero.

3. E between the plates is:

1. 17 × 10^-22 N/C
2. 1.5 × 10^-15 N/C
3. 1.9 × 10^-10 N/C
4. Zero.

4. The ratio of E from right side of B at distances 2cm and 4cm, respectively is:

1. 1 : 2
2. 2 : 1
3. 1 : 1
4. $1:\sqrt{2}$

5. ln order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is:

1. Spherical.
2. Spherical.
3. Straight line.
4. None of these.