Question:
A small ball of mass 5 g is suspended by a string of length 20 cm in a horizontal uniform electric field of \(1 \times 10^3~\text{N/C}\). If the ball is in equilibrium when the string makes an angle 60° with the vertical, then the net charge on the ball is:
[Acceleration due to gravity = 10 m/s\(^2\)]
A small ball of mass 5 g is suspended by a string of length 20 cm in a horizontal uniform electric field of \(1 \times 10^3~\text{N/C}\). If the ball is in equilibrium when the string makes an angle 60° with the vertical, then the net charge on the ball is:
[Acceleration due to gravity = 10 m/s\(^2\)]
[Acceleration due to gravity = 10 m/s\(^2\)]
Show Hint
For a charged mass in equilibrium in an electric field, resolve tension into vertical (balances weight) and horizontal (balances electric force) components to solve for charge.
Updated On: Jun 19, 2026
- 52.5 \(\mu\)C
- 46.4 \(\mu\)C
- 96.2 \(\mu\)C
- 86.6 \(\mu\)C
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The Correct Option is D
Solution and Explanation
Step 1: Resolve forces on the ball.
- Gravitational force: \(F_g = mg = 0.005 \cdot 10 = 0.05~\text{N}\)
- Electric force: \(F_e = q E\) along horizontal
- Tension T along the string, angle \(\theta = 60^\circ\) with vertical.
Step 2: Equilibrium conditions.
\[ \text{Vertical: } T \cos \theta = F_g \Rightarrow T = \frac{F_g}{\cos 60^\circ} = \frac{0.05}{0.5} = 0.1~\text{N} \] \[ \text{Horizontal: } T \sin \theta = F_e \Rightarrow F_e = 0.1 \cdot \sin 60^\circ = 0.1 \cdot 0.866 = 0.0866~\text{N} \]
Step 3: Find charge.
\[ F_e = q E \Rightarrow q = \frac{F_e}{E} = \frac{0.0866}{1000} = 8.66 \times 10^{-5}~\text{C} = 86.6~\mu\text{C} \]
Step 4: Conclusion.
The net charge on the ball is 86.6 \(\mu\)C.
- Gravitational force: \(F_g = mg = 0.005 \cdot 10 = 0.05~\text{N}\)
- Electric force: \(F_e = q E\) along horizontal
- Tension T along the string, angle \(\theta = 60^\circ\) with vertical.
Step 2: Equilibrium conditions.
\[ \text{Vertical: } T \cos \theta = F_g \Rightarrow T = \frac{F_g}{\cos 60^\circ} = \frac{0.05}{0.5} = 0.1~\text{N} \] \[ \text{Horizontal: } T \sin \theta = F_e \Rightarrow F_e = 0.1 \cdot \sin 60^\circ = 0.1 \cdot 0.866 = 0.0866~\text{N} \]
Step 3: Find charge.
\[ F_e = q E \Rightarrow q = \frac{F_e}{E} = \frac{0.0866}{1000} = 8.66 \times 10^{-5}~\text{C} = 86.6~\mu\text{C} \]
Step 4: Conclusion.
The net charge on the ball is 86.6 \(\mu\)C.
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