Question

Two point charges Q and \( -q \) are held \( r \) distance apart in free space. A uniform electric field \( \vec{E} \) is applied in the region perpendicular to the line joining the two charges. Which one of the following angles will the direction of the net force acting on charge \( -q \) make with the line joining Q and \( -q \)?

• A. \( \tan^{-1} \left( \frac{4\pi \epsilon_0 E r^2}{Q} \right) \)
• B. \( \cot^{-1} \left( \frac{4\pi \epsilon_0 E r^2}{Q} \right) \)
• C. \( \tan^{-1} \left( \frac{QE}{4\pi \epsilon_0 r^2} \right) \)
• D. \( \cot^{-1} \left( \frac{QE}{4\pi \epsilon_0 r^2} \right) \)

Hint:

The angle of the net force on charge \( -q \) is found by comparing the magnitudes of the forces due to the electric field and Coulomb's law. Use the tangent of the angle to find the result.

Answer 1

The Correct Option is A

The point charges \( Q \) and \( -q \) are subjected to forces from each other and an applied electric field \( \vec{E} \). The force exerted by the electric field on charge \( -q \) is \( F_{\text{field}} = qE \), acting perpendicular to the line connecting the charges. Coulomb's law describes the force between the two charges: \( F_{\text{Coulomb}} = \frac{1}{4\pi \epsilon_0} \frac{Qq}{r^2} \). The net force on \( -q \) is the vector sum of these two forces. If \( \theta \) is the angle between the line joining the charges and the net force on \( -q \), then \( \tan \theta = \frac{F_{\text{field}}}{F_{\text{Coulomb}}} \). Substituting the force expressions yields \( \tan \theta = \frac{qE}{\frac{1}{4\pi \epsilon_0} \frac{Qq}{r^2}} \), which simplifies to \( \tan \theta = \frac{4\pi \epsilon_0 E r^2}{Q} \). Therefore, the angle \( \theta \) is given by \( \theta = \tan^{-1} \left( \frac{4\pi \epsilon_0 E r^2}{Q} \right) \).