Question

Two point charges $q_1$ and $q_2$ having charge$q_1=+16\mu \text{C}$ and $q_2=+4\mu \text{C}$, are separated in vacuum by a distance of $3.0\text{m}$. Find the point on the line joining these charges where the net electric field is zero.

• A. $2\text{m}$ from $q_1$
• B. $1\text{m}$ from $q_1$
• C. $2\text{m}$ from $q_2$
• D. $1.5\text{m}$ from $q_2$

Answer 1

The correct option is A $2\text{m}$ from $q_1$

Consider the charges $q_1$ and $q_2$ separated by a distance of $3.0\text{m}$. Let the electric field due to charge $q_1$ be $E_1$ and the electric field due to charge $q_2$ be $E_2$.


We can see that between the charges, the two field contributions have opposite directions. So we can consider a point $P$ where the net electric field is zero. At point $P$ the magnitudes of $E_1$ and $E_2$ are equal.


However, since $q_2<q_1$, point $P$ must be closer to $q_2$, so that the field of the smaller charge can balance the field of the larger charge.

At $P,E_1=E_2$

$\because E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

we have, $\frac{1}{4\pi {ϵ}_0}\frac{q_1}{r_1^2}=\frac{1}{4\pi {ϵ}_0}.\frac{q_2}{r_2^2}$

$\therefore \frac{r_1}{r_2}=\sqrt{\frac{q_1}{q_2}}=\sqrt{\frac{16}{4}}=2$

$\Rightarrow r_1=2r_2\dots \dots \left(i\right)$

Also, $r_1+r_2=3.0\text{m}\dots \dots \left(ii\right)$

Solving these equations, we get

$2r_2+r_2=3\text{m}$

$\Rightarrow r_2=1\text{m}$

$\Rightarrow r_1=2r_2=2\text{m}$

Thus, we can say that the point $P$ is at a distance of $2\text{m}$ from $q_1$ and $1\text{m}$ from $q_2$.

Hence, option (a) is the correct answer.