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, \[\sigma =26.4\times {{10}^{-12}}\,C/{{m}^{2}},\]              \[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}C/N-{{m}^{2}}\] 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.