Rankers Physics
Topic: Electrostatics
Subtopic: Gauss's Law

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude 27\times 10^{-22}Cm^{-2}. The electric field \overrightarrow{E} in region II in between the plates is : Image related to
\[ 4.25\times 10^{-8}NC^{-1}\]
\[ 6.28\times 10^{-10}NC^{-1}\]
\[ 3.05\times 10^{-10}NC^{-1}\]
\[ 5.03\times 10^{-10}NC^{-1}\]

Solution:

Electric field at Point II will be due to both the sheets so,

\[\overrightarrow{E}=\frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}\]

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