Question

Two point charges $q$ and $9q$ are placed at distance of $l$ from each other. Then the electric field is zero at a

• A. Distance $\frac{l}{4}$ from charge $9q$
• B. Distance $\frac{3l}{4}$ from charge $q$
• C. Distance $\frac{l}{3}$ from charge $9q$
• D. Distance $\frac{l}{4}$ from charge $q$

Answer 1

Answer: (D)

Distance $\frac{l}{4}$ from charge $q$

Answer 2

Let the charges $q$ and $9q$ be at $x=0$ and $x=l$, respectively. A point $x$ (with $0 < x < l$) experiences:

• Field from $q$: $E_q = \dfrac{kq}{x^2}$ (to the right).

• Field from $9q$: $E_{9q} = \dfrac{k(9q)}{(l-x)^2}$ (to the left).

For net $E=0$, equate the magnitudes:

$$\dfrac{kq}{x^2} = \dfrac{k(9q)}{(l-x)^2}.$$

Cancel common factors:

$$\dfrac{1}{x^2} = \dfrac{9}{(l-x)^2} \quad \Longrightarrow \quad (l-x)^2 = 9x^2.$$

Taking the square root:

$$l-x = 3x \quad \Longrightarrow \quad l = 4x \quad \Longrightarrow \quad x = \frac{l}{4}.$$

Thus, the zero field point is at a distance $\frac{l}{4}$ from charge $q$.