Question

Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capacitance of the first capacitor is C, then capacitance of the second one is:

Answer 1

Answer:

Capacitance of the second capacitor = 2C

Explanation:

$\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}$

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have    

              $\texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C$

Similarly for capacitor 2

               $\texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C$

Capacitance of the second capacitor = 2C