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)  \[\text{1}\text{.5}\,\text{N/C}\]                            

B)  \[\text{1}\text{.5 }\!\!\times\!\!\text{ }\,\text{1}{{\text{0}}^{\text{-10}}}\,\text{N/C}\]

C)  \[3\,\text{N/C}\]                           

D)  \[\text{3 }\!\!\times\!\!\text{ }\,\text{1}{{\text{0}}^{\text{-10}}}\text{N/C}\]

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,   \[\sigma =2.64\times {{10}^{-12}}C/{{m}^{2}},\]                 \[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}/N\text{-}{{m}^{2}}\] Hence,  \[E=\frac{26.4\times {{10}^{-12}}}{8.85\times {{10}^{-12}}}\approx 3\,\,N/C\]