question_answer

Two parallel large thin metal sheets have equal surface charge densities \[(\sigma =26.4\times {{10}^{-12}}C/{{m}^{2}})\] of opposite signs. The electric field between these sheets is:

A) \[1.5\,\,N/C\]                   

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

C)  \[3\,\,N/C\]                      

D)  \[3\times {{10}^{-10}}N/C\]

Correct Answer: C

Solution :

The situation is shown in the figure. Plate \[1\] has surface charge density \[\sigma \] and plate \[2\] has surface charge density \[-\sigma \]. The electric field at point \[P\] due to two charged plates add up, giving                 \[E=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}=\frac{\sigma }{{{\varepsilon }_{0}}}\] Given,   \[\sigma =26.4\times {{10}^{-12}}C/{{m}^{2}}\]                 \[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}C/N\text{-}{{m}^{2}}\] Hence,   \[E=\frac{26.4\times {{10}^{-12}}}{8.85\times {{10}^{-12}}}\approx 3\,\,N/C\] Note: The direction of electric field is from the positive to the negative plate.