Question

Three point charges \(+4e\) , \(+2e\) and \(+e\) are kept at each vertex of an equilateral triangle ABC , each of side \(a\) . The net electric intensity acting on charge \(+e\) is
( \(\cos 60^{\circ }=1/2\) and \(k\) is constant )

• A. \(\frac{2\sqrt{7}ke}{a^{2}}\)
• B. \(\frac{4\sqrt{7}ke}{a^{2}}\)
• C. \(\frac{7\sqrt{2}ke}{a^{2}}\)
• D. \(\frac{\sqrt{7}ke}{a^{2}}\)

Hint:

Find the field due to each of the other two charges at the vertex and add them as vectors with a 60 degree angle between them.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The field at the vertex holding \(+e\) is due to the other two charges, each at distance \(a\). Both point away from their source charges.

Step 2: Magnitudes
\(E_{1}=\frac{k(4e)}{a^{2}}=\frac{4ke}{a^{2}}\) and \(E_{2}=\frac{k(2e)}{a^{2}}=\frac{2ke}{a^{2}}\). The angle between them is \(60^{\circ}\).

Step 3: Resultant
\(E=\sqrt{E_{1}^{2}+E_{2}^{2}+2E_{1}E_{2}\cos60^{\circ}}=\frac{ke}{a^{2}}\sqrt{16+4+8}=\frac{ke}{a^{2}}\sqrt{28}=\frac{2\sqrt{7}ke}{a^{2}}\).

Final Answer:
The net field is 2 root 7 ke divided by a squared. \[ \boxed{\text{(A) }\dfrac{2\sqrt{7}ke}{a^{2}}} \]