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 o and plate 2 has surface charge density \[-\sigma .\] The electric fields at point P due to two charged plates add up, giving \[E=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}=\frac{\sigma }{{{\varepsilon }_{0}}}\] 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.