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 fields at point P due to two charged plates add up, giving                 \[E=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2\varepsilon }=\frac{\sigma }{{{\varepsilon }_{{}}}}\] Given,   \[\sigma =26.4\times {{10}^{-12}}C/{{m}^{2}}\],                 \[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}/N-{{m}^{2}}\]. 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.