Question

The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

• A. \( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
• B. \( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
• C. \( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
• D. Zero

Hint:

For a system of charges with symmetry, use Coulomb’s Law to calculate the electric field, and remember that fields due to symmetric charges may cancel out.

Answer 1

The Correct Option is C

We have three charges at the vertices of a triangle ABC. The charge at vertex A is \( 2q \), and the charges at vertices B and C are \( -q \). The net electric field at the midpoint M of the base BC is the vector sum of the electric fields due to each charge. - The electric field due to the charge at B and C will cancel each other out because of symmetry.
- The electric field due to the charge at A will point along the line AM (because the distance from A to M is along the line connecting A and M).
The magnitude of the electric field at M due to the charge at A is given by Coulomb's Law: \[ E = \frac{1}{4 \pi \epsilon_0} \frac{2q}{l^2} \] Thus, the net electric field is: \[ E = \frac{q}{2 \pi \epsilon_0 l^2} \text{ pointing along AM} \] Final Answer: (C) \( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM