Question

The distance between the plates of a parallel plate capacitor is $d$. A metal plate of thickness $d/2$ is placed between the plates. The capacitance would be then be

• A. Unchanged
• B. Initial
• C. Zero
• D. Doubled

Answer 1

Answer: (B)

$\begin{array}{c}CaseI:\left(Airbetweenplates\right)\\ Capaci\tan ce,\\ C=\frac{E_{0}A}{d}\\ CaseII:\left(Metalplateisinsertedbetweentheplates\right)\\ capaci\tan e,\\ C^{\prime }=\frac{E_{0}A}{d-t+\frac{t}{k}}\\ Where\\ t=thicknessofmetalplate=\frac{d}{2}\\ k=Dielectriccons\tan tofmetalplate=\infty ]\\ C^{\prime \prime }=\frac{E_{0}A}{d-t+\frac{\frac{d}{2}}{\infty }}\\ =\frac{E_{0}A}{\frac{d}{2}}\\ =\frac{2E_{0}A}{d}\\ =2C\\ Hence,itcanbeconculedthatthenewcapaci\tan cebecomesdoublethatofinitialcapaci\tan e.\end{array}$