Question

Two point charges of magnitudes \( -8 \mu C \) and \( +32 \mu C \) are separated by a distance of 15 cm in air. The position of the point from the \( -8 \mu C \) charge at which the resultant electric field becomes zero is:

• A. \( 15 \) cm
• B. \( 30 \) cm
• C. \( 7.5 \) cm
• D. \( 5 \) cm

Hint:

The point where the net electric field is zero lies outside the two charges when the charges are opposite in sign.
- Use the inverse square law \( E = \frac{k q}{r^2} \) to equate the electric fields.

Answer 1

The Correct Option is A

The electric field due to a point charge is given by: \[ E = \frac{k q}{r^2} \] where: - \( q \) is the charge, - \( r \) is the distance from the charge, - \( k \) is Coulomb's constant. Let the required point be at a distance \( x \) from \( -8 \mu C \). 1. Condition for zero electric field: The net electric field is zero where the fields due to both charges cancel out. Since the positive charge is larger, the point must be outside the two charges. 2. Equating electric fields: \[ \frac{k (8)}{x^2} = \frac{k (32)}{(15+x)^2} \] Simplifying, \[ \frac{8}{x^2} = \frac{32}{(15+x)^2} \] \[ \frac{1}{x^2} = \frac{4}{(15+x)^2} \] Taking square roots, \[ \frac{1}{x} = \frac{2}{15+x} \] \[ 15+x = 2x \] \[ x = 15 \text{ cm} \] Thus, the correct answer is \(\boxed{15 \text{ cm}}\).