• # 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)  $3\times {{10}^{-10}}N/C$

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.