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:
For two charges of opposite signs, the null point (where the net electric field is zero) lies outside the line segment joining the charges, on the side of the smaller magnitude charge.
Step 2: Key Formula or Approach:
The magnitude of electric field due to a point charge is \( E = \frac{kq}{r^2} \).
Let the null point be at a distance \( x \) from the smaller charge \( q_2 \).
The distance from the larger charge \( q_1 \) will be \( (d + x) \), where \( d = 30 \, \text{cm} \).
At the null point: \( E_1 = E_2 \).
Step 3: Detailed Explanation:
Equating the fields:
\[ \frac{k |q_1|}{(d + x)^2} = \frac{k |q_2|}{x^2} \]
Substituting values:
\[ \frac{36}{(30 + x)^2} = \frac{9}{x^2} \]
Taking the square root of both sides:
\[ \frac{6}{30 + x} = \frac{3}{x} \]
Cross-multiply:
\[ 6x = 3(30 + x) \]
\[ 6x = 90 + 3x \]
\[ 3x = 90 \implies x = 30 \, \text{cm} \]
The question asks for the distance from \( q_1 \).
Distance from \( q_1 = d + x = 30 + 30 = 60 \, \text{cm} \).
Step 4: Final Answer:
The distance from \( q_1 \) where the net electric field is zero is 60 cm.