WBJEE · Physics · Electrostatics
Two infinite parallel metal planes, contain electric charges with charge densities \(+\sigma\) and \(-\sigma\) respectively and they are separated by a small distance in air. If the permittivity of air is \(\varepsilon_{0}\), then the magnitude of the field between the two planes with its direction will be
- A \(\sigma / \varepsilon_{0}\) towards the positively charged plane
- B \(\sigma / \varepsilon_{0}\) towards the negatively charged plane
- C \(\sigma /\left(2 \varepsilon_{0}\right)\) towards the positively charged plane
- D 0 and towards any direction
Answer & Solution
Correct Answer
(B) \(\sigma / \varepsilon_{0}\) towards the negatively charged plane
Step-by-step Solution
Detailed explanation
\(E=\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}\) \(E=\frac{\sigma}{\varepsilon_{0}}\) towards negative charged plate.
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