Question

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A has a radius 2 times that of sphere B. Let QA and QB be the charges on the two spheres, and let EA and EB be the electric-field magnitudes at the surfaces of the two spheres.What is the ratio QB/QA?What is the ratio EB/EA?

Answer 1

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

$R_A=2R_B$

Also Potential of both sphere is same

$V_A=V_B$

$V=\frac{kQ}{R}$

thus

$k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}$

$\frac{Q_A}{Q_B}=\frac{R_A}{R_B}$

$\frac{Q_A}{Q_B}=\frac{2}{1}=2$

$\frac{Q_B}{Q_A}=\frac{1}{2}$

(b)Ratio of $\frac{E_B}{E_A}$

Electric Field is given by $E=\frac{kQ}{R^2}$

thus $E_A=\frac{kQ_A}{R_A^2}$----1

$E_B=\frac{kQ_B}{R_B^2}$----2

Divide 2 by 1

$\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}$

$\frac{E_B}{E_A}=\frac{1}{2}\times 4=2$