Question
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 $17.0 \times 10^{-22}Cm^{-2}$ 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. $17 \times 10^{-22} N/C$
  2. $1.5 \times 10^{-25} N/C$
  3. $1.9 \times 10^{-10} N/C$
  4. Zero.
  1. E in the outer region of the second plate is:
  1. $17 \times 10^{-22} N/C$
  2. $1.5 \times 10^{-15} N/C$
  3. $1.9 \times 10^{-10} N/C$
  4. Zero.
  1. E between the plates is:
  1. $17 \times 10^{-22} N/C$
  2. $1.5 \times 10^{-15} N/C$
  3. $1.9 \times 10^{-10} N/C$
  4. Zero.
  1. 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. $1:\sqrt{2}$
  1. 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.

Answer

  1. (d) Zero.
Explanation:
There are two plates A and B having surface charge densities,

$\sigma_\text{A}=17.0\times10^{-22}\text{C/m}^2$
on A and $\sigma_\text{B}=-17.0\times10^{-22}\text{C/m}^2$on B, respectively. According to Gauss' theorem, if the plates have same surface charge density but having opposite signs, then the electric field in region I is zero.
$\text{E}_\text{I}=\text{E}_\text{A}+\text{E}_\text{B}$
$=\frac{\sigma}{2\in_0}+\Big(-\frac{\sigma}{2\in_0}\Big)=0$
  1. (d) Zero.
Explanation:
The electric field in region III is also zero.
$\text{E}_\text{III}=\text{E}_\text{A}+\text{E}_\text{B}$
$=\frac{\sigma}{2\in_0}+\Big(-\frac{\sigma}{2\in_0}\Big)=0$
  1. (c)$ 1.9 \times 10^{-10} N/C$
​​​​​​​​​​​​​​​​​​​​​​​ Explanation:
In region II or between the plates, the electric field.
$\text{E}_\text{II}=\text{E}_\text{A}-\text{E}_\text{B}$
$=\frac{\sigma}{2\in_0}+=\frac{\sigma}{2\in_0}$
$=\frac{\sigma(\sigma_\text{A}\text{ or }\sigma_\text{B})}{\in_0}=\frac{17.0\times10^{-22}}{8.85\times10^{-12}}$
$E = 1.9 \times 10^{-10} N/C$​​​​​​​
  1. (c) 1 : 1
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​ Explanation:
Since electric field due to an infinite-plane sheet of charge does not depend on the distance of observation point from the plane sheet of charge. So, for the given distances, the ratio of E will be 1 : 1.
  1. (b) Spherical.
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​ Explanation:
ln order to estimate the electric field due to a thin finite plane metal plate, we take a cylindrical cross-sectional area A and length 2r as the gaussian surface.

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