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: Use the formula for electric field due to a point charge:
The electric field due to each point charge is given by: \[ E = \frac{k|Q|}{r^2} \] For the electric field to be zero, the fields due to the two charges must cancel each other out.
Step 2: Apply the condition for zero electric field: Let the distance from the \( +4 \mu C \) charge be \( x \) and the distance from the \( -2 \mu C \) charge be \( 1 - x \). The magnitudes of the electric fields must be equal, so: \[ \frac{k(4 \times 10^{-6})}{x^2} = \frac{k(2 \times 10^{-6})}{(1 - x)^2} \]
Step 3: Solve for \( x \): \[ \frac{4}{x^2} = \frac{2}{(1 - x)^2} \] Solving this equation gives: \[ x = 0.58 \, \text{m} \]