A point charge $$q_1 = 4q_0$$ is placed at origin. Another point charge $$q_2 = -q_0$$ is placed at $$x = 12$$ cm. Charge of proton is $$q_0$$. The proton is placed on $$x$$-axis so that the electrostatic force on the proton in zero. In this situation, the position of the proton from the origin is ______ cm.

Solution :

Given :

$$q_1 = 4q_0$$ at origin

$$q_2 = -q_0$$ at $$x = 12\text{ cm}$$

Let the proton be placed at a distance $$x$$ from the origin where net electrostatic force is zero.

Since the charges are opposite in sign, the zero force point cannot lie between them. It lies on the right side of $$q_2$$.

Let proton be at $$x > 12$$ cm.

Force due to $$q_1$$ on proton :

$$F_1 = \frac{k(4q_0)(q_0)}{x^2}$$

Force due to $$q_2$$ on proton :

$$F_2 = \frac{k(q_0)(q_0)}{(x-12)^2}$$

For net force to be zero :

$$F_1 = F_2$$

$$\frac{4}{x^2} = \frac{1}{(x-12)^2}$$

Taking square root :

$$\frac{2}{x} = \frac{1}{x-12}$$

$$2(x-12)=x$$

$$2x-24=x$$

$$x=24\text{ cm}$$

Final Answer :

$$24$$