Question

Two charged conducting spheres of radii a and b are connected to each other by a wire. The ratio of electric fields at the surfaces of two spheres is:

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $\frac{a_2}{b_2}$
• D. $\frac{b^2}{a^2}$

Answer 1

The correct option is B $\frac{b}{a}$

Let $a$ be the radius of a sphere $A$, $Q_a$ be the charge on the sphere, and $C_a$ be the capacitance of the sphere.

Let $b$ be the radius of a sphere $B$, $Q_b$ be the charge on the sphere, and $C_b$ be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential $\left(V\right)$ will become equal.

Let $E_a$ be the electric field of sphere A and $E_b$ be the electric field of sphere B. Therefore, their ratio,

$⟹\frac{E_a}{E_b}=\frac{Q_a}{4\pi {ϵ}_0{a}^2}\times \frac{4\pi {ϵ}_0{b}^2}{Q_b}$

$⟹\frac{E_a}{E_b}=\frac{Q_a}{Q_b}\times \frac{b^2}{a^2}$

However, $\frac{Q_a}{Q_b}=\frac{C_{a}V}{C_{b}V}$

And, $\frac{C_a}{C_b}=\frac{a}{b}$

$\therefore \frac{Q_a}{Q_b}=\frac{a}{b}$

$⟹\frac{E_a}{E_b}=\frac{a}{b}\frac{b^2}{a^2}=\frac{b}{a}$

Therefore, the ratio of electric fields at the surface is $\frac{b}{a}$