Question 651: Electrostatics & Capacitance

Question

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 is zero. In this situation, the position of the proton from the origin is ___________ cm.

Accepted Answer

Let a proton having charge $q_0$ on the $x$ axis at distance $x$ from $q_2$ and $(12+x)$ distance from $q_1$.
Now, balance the force between them $\vec{F}_1+\vec{F}_2=0$
$$
\frac{K 4 q_0\left(q_0ight)}{(12+x)^2}+\frac{K\left(-q_0ight)\left(q_0ight)}{x^2}=0 $$
$$ \Rightarrow $$ $$
 \frac{4 K q_0^2}{(12+x)^2}=\frac{K\left(q_0ight)^2}{x^2}
$$
$$ \Rightarrow $$ $$
\frac{4}{(12+x)^2}=\frac{1}{x^2} $$
$$ \Rightarrow $$ $$ \frac{2}{12+x}=\frac{1}{x} $$
$$ \Rightarrow $$ $$ 2 x=12+x $$
$$ \Rightarrow $$ $$ x=12 \mathrm{~cm}
$$
The charge $q_0$ is at distance of $24 \mathrm{~cm}$ from charge $q_1$ on $x$-axis.
$$ 	herefore $$ Distance from origin is $12+12=24 \mathrm{~cm}$

Electrostatics & Capacitance

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 is zero. In this situation, the position of the proton from the origin is ___________ cm.

Integer / Numeric Answer Type