Question

A parallel plate capacitor with circular plates and a capacitance $C$ has the radius of each plate doubled and the distance between the plates doubled.
What is the new capacitance?

• A. $\frac{C}{2}$
• B. $C$
• C. $2C$
• D. $4C$

Answer 1

The correct option is C $2C$

Area of the plates initially $A=\pi r^2$

Initial capacitance of a parallel plate capacitor $C=\frac{A{ϵ}_o}{d}=\frac{\left(\pi r^2\right){ϵ}_o}{d}$

Given : $d^{\prime }=2d$ $r^{\prime }=2r$

Thus new area of the plates $A^{\prime }=\pi (2r{)}^2=4\pi r^2$

$\therefore$ New capacitance $C^{\prime }=\frac{\left(4\pi r^2\right){ϵ}_o}{2d}=2C$