Question

Two-point charges, $q_1 = 36\ \mu\text{C}$ and $q_2 = -9\ \mu\text{C}$ are placed at a distance of $30\text{ cm}$. What distance from $q_1$, where the net electric field is zero, will be:

• A. $10\text{ cm}$
• B. $20\text{ cm}$
• C. $60\text{ cm}$
• D. $30\text{ cm}$

Hint:

For two point charges $q_1$ and $q_2$ separated by $r$, the distance of the neutral point from the smaller charge $q_2$ is $x = \frac{r}{\sqrt{q_1/q_2} \pm 1}$. Use $+$ if charges are opposite, $-$ if they are same.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The net electric field is zero at a point (the neutral point) where the field intensities of both charges are equal in magnitude and opposite in direction. Since the charges have opposite signs, the neutral point must lie outside the charges, on the side of the smaller magnitude charge ($q_2$).

Step 2: Identifying the Formula and Values:
Let $x$ be the distance of the neutral point from $q_2$. The distance from $q_1$ will be $(30 + x)$. \[ E_1 = E_2 \implies \frac{1}{4\pi\epsilon_0} \frac{|q_1|}{(30+x)^2} = \frac{1}{4\pi\epsilon_0} \frac{|q_2|}{x^2} \] \[ \frac{36}{(30+x)^2} = \frac{9}{x^2} \]

Step 3: Calculation:
Taking the square root on both sides: \[ \frac{6}{30+x} = \frac{3}{x} \] \[ 6x = 3(30 + x) \implies 6x = 90 + 3x \] \[ 3x = 90 \implies x = 30\text{ cm} \] The distance from $q_1$ is $30 + x = 30 + 30 = 60\text{ cm}$.

Step 4: Final Answer:
The neutral point is $60\text{ cm}$ away from $q_1$.