Question

Two point charges \( +4 \mu C \) and \( -2 \mu C \) are separated by a distance of 1 m. Then, the distance of the point on the line joining the charges, where the resultant electric field is zero, is (in metre):

• A. 0.58
• B. 0.75
• C. 0.67
• D. 0.81

Hint:

To find the point where the electric field is zero, equate the electric fields due to the two charges and solve for the distance.

Answer 1

The Correct Option is A

Step 1: Condition for zero electric field
Electric field due to a point charge is:
\[ E = \frac{k|Q|}{r^2} \] For zero net field, fields due to both charges must be equal in magnitude and opposite in direction.

Step 2: Setup equation
Let the point be at distance \(x\) from \(+4\mu C\), and \(1-x\) from \(-2\mu C\).
\[ \frac{k \cdot 4}{x^2} = \frac{k \cdot 2}{(1-x)^2} \] Cancel \(k\):
\[ \frac{4}{x^2} = \frac{2}{(1-x)^2} \]

Step 3: Simplify
\[ \frac{2}{x^2} = \frac{1}{(1-x)^2} \] Taking square root:
\[ \frac{\sqrt{2}}{x} = \frac{1}{1-x} \]

Step 4: Solve
\[ \sqrt{2}(1-x) = x \] \[ \sqrt{2} = x(1+\sqrt{2}) \] \[ x = \frac{\sqrt{2}}{1+\sqrt{2}} \] Rationalizing: \[ x \approx 0.586\ \text{m} \]

Final Answer:
\( x \approx 0.58\ \text{m} \)