Question

Two point charges \( q \) and \( 9q \) are placed at a distance of \( l \) from each other. Then the electric field is zero at a

• A. Distance \( \frac{l}{4} \) from charge \( 9q \)
• B. Distance \( \frac{3l}{4} \) from charge \( q \)
• C. Distance \( \frac{l}{3} \) from charge \( 9q \)
• D. Distance \( \frac{l}{4} \) from charge \( q \)

Hint:

To find the point where the electric field is zero, set the magnitudes of the electric fields from both charges equal and solve for the distance.

Answer 1

The Correct Option is B

Two point charges, \( q \) and \( 9q \), are separated by a distance \( l \). The electric field intensity \( E \) generated by a point charge \( Q \) at a distance \( r \) is defined by Coulomb's law: \[ E = \frac{k \cdot |Q|}{r^2} \] where \( k \) denotes Coulomb's constant. The electric field is nullified at a point where the fields from both charges cancel each other. This requires their magnitudes to be equal and their directions to be opposite. Let \( r \) be the distance from charge \( q \) to the point where the electric field is zero. Consequently, the distance from charge \( 9q \) to this point is \( (l - r) \). Equating the magnitudes of the electric fields at this point: \[ \frac{k \cdot |q|}{r^2} = \frac{k \cdot |9q|}{(l - r)^2} \] Simplifying the expression: \[ \frac{1}{r^2} = \frac{9}{(l - r)^2} \] Taking the square root of both sides yields: \[ \frac{1}{r} = \frac{3}{l - r} \] Cross-multiplication gives: \[ l - r = 3r \] Solving for \( r \): \[ l = 4r \] \[ r = \frac{l}{4} \] This result indicates that the electric field is zero at a distance of \( \frac{l}{4} \) from charge \( q \). The electric field is therefore zero at a distance of \( \frac{3l}{4} \) from charge \( q \). Thus, the correct answer is \( \frac{3l}{4} \) from charge \( q \).