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MHT CET · Physics · Capacitance

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. A \(\frac{\mathrm{Qt}}{\mathrm{C}}\)
  2. B \(\frac{\mathrm{Q}}{\mathrm{Ct}}\)
  3. C \(\frac{\mathrm{C}}{\mathrm{Qt}}\)
  4. D \(\frac{\mathrm{Ct}}{\mathrm{Q}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\mathrm{Q}}{\mathrm{Ct}}\)

Step-by-step Solution

Detailed explanation

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)]