Charge on a parallel plate capacitor of capacity C is Q, the electric field intensity between its two plates…

Charge on a parallel plate capacitor of capacity C is Q, the electric field intensity between its two plates separated by a distance of $t$ is
  1. $\frac{\mathrm{Qt}}{\mathrm{C}}$
  2. $\frac{\mathrm{Q}}{\mathrm{Ct}}$
  3. $\frac{\mathrm{C}}{\mathrm{Qt}}$
  4. $\frac{\mathrm{Ct}}{\mathrm{Q}}$

Solution

Electric field intensity between plates of a parallel plate capacitor is $\mathrm{E}=\frac{\mathrm{Q}}{\mathrm{~A} \varepsilon}$
Capacitance of a parallel plate capacitor is $\begin{aligned} & \quad \mathrm{C}=\frac{\mathrm{A} \varepsilon}{\mathrm{t}} \Rightarrow \mathrm{~A} \varepsilon=\mathrm{Ct} \\ & \therefore \quad \mathrm{E}=\frac{\mathrm{Q}}{\mathrm{Ct}} \end{aligned}$ ...[From(i) and (ii)]

Asked in: MHT CET 2024 (02 May Shift 2)

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