Electrostatics PYQ | Equilibrium of Charges | JEE Main Physics 2025

 

❓ Question

Two positive charges, $+4q$ and $+q$, are separated by a distance $r$.

Find the position on the line joining the charges where a test charge experiences zero net force.

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✍️ Solution

Let the equilibrium point be at a distance

$$x$$

from the charge $+4q$.

Then its distance from the charge $+q$ is

$$r-x.$$

Since both charges are positive, the equilibrium point lies between the two charges.

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Condition for Equilibrium

At equilibrium, the magnitudes of the forces due to both charges must be equal.

$$F_{4q}=F_q.$$

Using Coulomb's law,

$$\frac{k(4q)Q}{x^2} = \frac{k(q)Q}{(r-x)^2},$$

where $Q$ is the test charge.

Cancelling common terms,

$$\frac{4}{x^2} = \frac{1}{(r-x)^2}.$$

Cross-multiplying,

$$(r-x)^2=\frac{x^2}{4}.$$

Taking square roots,

$$r-x=\pm\frac{x}{2}.$$

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Case 1:

$$r-x=\frac{x}{2}$$
$$2r-2x=x$$
$$2r=3x$$
$$x=\frac{2r}{3}.$$

This satisfies

$$0<x<r,$$

so it is valid.

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Case 2:

$$r-x=-\frac{x}{2}$$
$$2r-2x=-x$$
$$2r=x,$$

which gives

$$x>r,$$

so this is not possible for a point between the charges.

Hence, this case is rejected.

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✅ Final Answer

The equilibrium point is

$$\boxed{\frac{2r}{3}}$$

from the charge $+4q$,

or equivalently,

$$\boxed{\frac{r}{3}}$$

from the charge $+q$.