question_answer

Direction: Q.26 to Q.30

Surface charge density is defined as charge per unit surface area of surface charge distribution, i.e.,\[\sigma =\frac{dq}{ds}\]. 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 \[17.0\times {{10}^{-22}}C{{m}^{-2}}\]as shown. The intensity of electric field at a point is\[E=\frac{\sigma }{{{\varepsilon }_{0}}}\],

where,\[{{\varepsilon }_{0}}\]= permittivity of free space.

Read the given passage carefully and give the answer of the following questions.

E in the outer region of the first plate is:

A) \[17\times {{10}^{-22}}N/C\]

B) \[1.5\times {{10}^{-15}}N/C\]

C) \[1.9\times {{10}^{-10}}N/C\]   

D) zero

Correct Answer: D

Solution :

(d) zero There are two plates A and B having surface charge densities, \[{{\sigma }_{A}}=17.0\times {{10}^{-22}}C/{{m}^{2}}\]on A and \[{{\sigma }_{B}}=-17.0\times {{10}^{-22}}\,C/{{m}^{2}}\]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. \[{{E}_{l}}={{E}_{A}}+{{E}_{B}}=\frac{\sigma }{2{{\varepsilon }_{0}}}+\left( \frac{\sigma }{2{{\varepsilon }_{0}}} \right)=0\]