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

We are given two point charges \( q \) and \( 9q \) placed at a distance \( l \) from each other. The electric field at a point due to a point charge is given by Coulomb's law: \[ E = \frac{k \cdot |Q|}{r^2} \] where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge. For the electric field to be zero at some point, the electric fields due to both charges must cancel each other out. That is, the magnitudes of the fields due to \( q \) and \( 9q \) must be equal, but they should be in opposite directions. Let the point where the electric field is zero be at a distance \( r \) from the charge \( q \), and \( (l - r) \) from the charge \( 9q \). At this point, the electric fields due to both charges should be equal in magnitude, so: \[ \frac{k \cdot |q|}{r^2} = \frac{k \cdot |9q|}{(l - r)^2} \] Simplifying the equation: \[ \frac{1}{r^2} = \frac{9}{(l - r)^2} \] Taking the square root of both sides: \[ \frac{1}{r} = \frac{3}{l - r} \] Cross-multiply: \[ l - r = 3r \] Solving for \( r \): \[ l = 4r \] \[ r = \frac{l}{4} \] Therefore, the electric field is zero at a distance \( \frac{3l}{4} \) from charge \( q \). Thus, the correct answer is \( \frac{3l}{4} \) from charge \( q \).