Question

The distance between charges $5\times 10^{-11}\mathrm{C}$ and $-2.7\times 10^{-11}\mathrm{C}$ is $0.2\mathrm{m}$. The distance at which a third charge should be placed from the second charge in order that it will not experience any force along the line joining the two charges is

• A. $0.44\mathrm{m}$
• B. $0.65\mathrm{m}$
• C. $0.556\mathrm{m}$
• D. $0.350\mathrm{m}$

Hint:

For zero net force, the forces from both charges must be equal in magnitude and opposite in direction.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For net force to be zero, forces from both charges must be equal and opposite.
Step 2: Detailed Explanation:
Let the third charge be placed at a distance $x$ from the second charge. $\frac{1}{4\pi\epsilon_0} \frac{5 \times 10^{-11} q}{x^2} = \frac{1}{4\pi\epsilon_0} \frac{2.7 \times 10^{-11} q}{(0.2 - x)^2}$. Solving gives $x = 0.556$ m from the second charge.
Step 3: Final Answer:
The distance is $0.556$ m.