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 \( -q \) experience forces due to each other and the applied electric field \( \vec{E} \). The force due to the electric field on charge \( -q \) is given by: \[ F_{\text{field}} = qE \] This force acts perpendicular to the line joining the charges. The force between the two charges is given by Coulomb's law: \[ F_{\text{Coulomb}} = \frac{1}{4\pi \epsilon_0} \frac{Qq}{r^2} \] The net force on charge \( -q \) will be the resultant of the force due to the electric field and the force due to Coulomb's interaction between the charges. Let \( \theta \) be the angle between the line joining the charges and the net force on charge \( -q \). Then, from the vector diagram, we can find the tangent of the angle \( \theta \): \[ \tan \theta = \frac{F_{\text{field}}}{F_{\text{Coulomb}}} = \frac{qE}{\frac{1}{4\pi \epsilon_0} \frac{Qq}{r^2}} \] Simplifying the equation: \[ \tan \theta = \frac{4\pi \epsilon_0 E r^2}{Q} \] Thus, the angle \( \theta \) is: \[ \theta = \tan^{-1} \left( \frac{4\pi \epsilon_0 E r^2}{Q} \right) \]