Question

Two identical metal balls with charges +2Q and -Q are separated by the same distance, and exert a force F on each other. They are joined by a conducting wire, which is then removed. The magnitude of the force between them will now be

Answer 1

Step 1: Given that:

The charge on the first metallic ball = $+2Q$

Charge on second metallic ball = $-Q$

The force exerted by both on each other = F

Step 2: Formula used:

According to Coulomb's law;

The force between two point charges separated by a small distance is given by;

$F=k_e\frac{q_1{q}_2}{r^2}$

Where $q_1$ and $q_2$ are the two charges and r is the distance between them.

When two charged bodies are brought in contact, the charges are distributed on the bodies in such a way that the charges on both bodies become equal.

Step 3: calculation of the force F between the given charges:

$F=k_e\frac{2Q\times Q}{r^2}$

$F=k_e\frac{2Q^2}{r^2}$..........(1)

Step 4: Calculation of the force between the charged bodies when they are brought in contact:

When both the bodies are borough in contact, the charge on each will be;

$q_1=q_2=\frac{+2Q-Q}{2}=\frac{Q}{2}$

Therefore the force F' between them will be;

$F^{\prime }=k_e\frac{\frac{Q}{2}\times \frac{Q}{2}}{r^2}$

$F^{\prime }=k_e\frac{Q^2}{4r^2}$ ..........(2)

Now,

By dividing equation 2) from equation 1) we get

$\frac{F^{\prime }}{F}=\frac{k_e\frac{Q^2}{4r^2}}{k_e\frac{2Q^2}{r^2}}$

$\frac{F^{\prime }}{F}=k_e\frac{Q^2}{4r^2}\times \frac{r^2}{k_{e}2Q^2}$

$\frac{F^{\prime }}{F}=\frac{1}{8}$

$F^{\prime }=\frac{F}{8}$

Thus,

The magnitude of the force will now be $\frac{F}{8}$ .