Question

Two point charges $ +Q $ and $ -4Q $ are placed 60 cm apart. Where should a third charge be placed so that it experiences zero net force?

• A. 15 cm from \( +Q \)
• B. 20 cm from \( -4Q \)
• C. 60 cm from \( +Q \)
• D. 45 cm from \( -4Q \)

Hint:

Always consider direction of forces when solving electrostatics problems. For equilibrium points, test regions outside the segment first, especially when dealing with unlike charges. Use Coulomb's law and set magnitudes equal to find the exact distance.

Answer 1

The Correct Option is C

Two charges, \( +Q \) and \( -4Q \), are separated by 60 cm. We seek a position for a third charge, \( q \), where the net electrostatic force on it is zero. The possible regions for placing \( q \) are: (i) to the left of \( +Q \), (ii) between \( +Q \) and \( -4Q \), and (iii) to the right of \( -4Q \).

Case (ii): Between \( +Q \) and \( -4Q \)
If \( q \) is placed between the charges, the forces exerted by \( +Q \) and \( -4Q \) will act in the same direction (either both attractive or both repulsive, depending on \( q \)'s sign). Therefore, cancellation of forces is not possible, and equilibrium cannot be achieved in this region.

Case (i): To the left of \( +Q \) 
Assume \( q \) is placed at a distance \( x \) cm to the left of \( +Q \). Its distance from \( -4Q \) will then be \( x + 60 \) cm. Applying Coulomb's law and setting the magnitudes of the forces equal for zero net force: \[ \frac{Q}{x^2} = \frac{4Q}{(x + 60)^2} \Rightarrow \frac{1}{x^2} = \frac{4}{(x + 60)^2} \Rightarrow \frac{(x + 60)^2}{x^2} = 4 \Rightarrow \frac{x + 60}{x} = 2 \Rightarrow x + 60 = 2x \Rightarrow x = 60 \text{ cm} \]

Thus, the third charge should be positioned 60 cm to the left of \( +Q \). This aligns with Option (C), indicating a point 60 cm from \( +Q \). Let's verify the force directions:
- The force exerted by \( +Q \) on \( q \) is repulsive.
- The force exerted by \( -4Q \) on \( q \) is attractive.
When \( q \) is to the left of \( +Q \), these forces act in opposite directions, allowing for cancellation, and thus equilibrium, only in this region.