Question

Two particles A and B having charges \(20\,\mu C\) and \(-5\,\mu C\) respectively are held fixed with a separation of \(5\,\text{cm}\). At what position a third charged particle should be placed so that it does not experience a net electric force ?

• A. At midpoint between two charges
• B. At \(5\,\text{cm}\) from \(-5\,\mu C\) on the right side
• C. At \(5\,\text{cm}\) from \(20\,\mu C\) on the left side of system
• D. At \(1.25\,\text{cm}\) from a \(-5\,\mu C\) between two charges

Hint:

For like charges, equilibrium is in between. For unlike charges, equilibrium is outside, closer to the smaller charge.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a third charge to experience zero net force (equilibrium) when placed near two unlike charges, it must be placed outside the segment joining them, on the side of the smaller magnitude charge.
Step 2: Key Formula or Approach:
Let the third charge \( q \) be placed at distance \( x \) from the charge \( q_2 = -5\,\mu C \).
The distance from \( q_1 = 20\,\mu C \) will be \( (5 + x)\,\text{cm} \).
Equate the magnitudes of forces: \( \frac{k |q_1| |q|}{(5+x)^2} = \frac{k |q_2| |q|}{x^2} \).
Step 3: Detailed Explanation:
\[ \frac{20}{(5+x)^2} = \frac{5}{x^2} \]
Divide both sides by 5:
\[ \frac{4}{(5+x)^2} = \frac{1}{x^2} \]
Take the square root of both sides:
\[ \frac{2}{5+x} = \frac{1}{x} \]
\[ 2x = 5 + x \]
\[ x = 5\,\text{cm} \]
The position is \( 5\,\text{cm} \) away from the \(-5\,\mu C\) charge, on the side away from the \(20\,\mu C\) charge (right side).
Step 4: Final Answer:
The third charge should be placed at \(5\,\text{cm}\) from \(-5\,\mu C\) on the right side.