Question

A hollow metal sphere, Sphere A, sits on an insulating stand. Sphere A has a diameter of 4 inches, and a net charge of magnitude $Q_0$. A second hollow metal sphere, Sphere B, also sits on an insulating stand, but has a diameter of 8 inches and zero net charge. The two spheres are brought close so that they touch, then they are separated.
In terms of $Q_0$, what is the final charge on Sphere A?

• A. $\frac{Q_0}{5}$
• B. $\frac{Q_0}{4}$
• C. $\frac{Q_0}{2}$
• D. $Q_0$
• E. $4Q_0$

Answer 1

The correct option is A $\frac{Q_0}{5}$

$R_B=2R_A$

Let final charge on spheres be $Q_A$ and $Q_B$.

On touching the spheres charge density becomes same.

And, charge density, $\sigma =\frac{Q}{4\pi R^2}$

$⟹Q\propto R^2$ for same $\sigma$

Hence, $\frac{Q_B}{Q_A}=\frac{R_B^2}{R_A^2}=2^2$

$⟹Q_B=4Q_A$

And, total charge$=Q_B+Q_A=Q_o$

$⟹4Q_A+Q_A=Q_o$

$⟹Q_A=\frac{Q_o}{5}$

Answer-(A)