Question:
A small sphere of charge \(50\,\mu C\) and mass \(5\,g\) is attached to a horizontal light string and placed in a uniform electric field which makes an angle \(30^\circ\) with the horizontal. The opposite end of the string is attached to a vertical wall. If the sphere is in static equilibrium and the string is horizontal, then the tension in the string is: (Take \(g = 10\,m\,s^{-2}\))
A small sphere of charge \(50\,\mu C\) and mass \(5\,g\) is attached to a horizontal light string and placed in a uniform electric field which makes an angle \(30^\circ\) with the horizontal. The opposite end of the string is attached to a vertical wall. If the sphere is in static equilibrium and the string is horizontal, then the tension in the string is: (Take \(g = 10\,m\,s^{-2}\))
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In equilibrium problems, always resolve forces along independent axes and first eliminate unknown field magnitude using one direction.
Updated On: Jun 19, 2026
- \(5.75 \times 10^{-2}\,N\)
- \(6.65 \times 10^{-2}\,N\)
- \(8.66 \times 10^{-2}\,N\)
- \(0.12\,N\)
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The Correct Option is C
Solution and Explanation
Step 1: Identify forces acting on the sphere.
The forces are: - Weight \(mg\) acting downward - Electric force \(qE\) acting at \(30^\circ\) above horizontal - Tension \(T\) acting horizontally (string is horizontal)
Step 2: Resolve electric force into components.
\[ (qE)_x = qE\cos 30^\circ,\quad (qE)_y = qE\sin 30^\circ \]
Step 3: Apply vertical equilibrium condition.
Since the string is horizontal, vertical forces balance only between weight and vertical electric component: \[ qE\sin 30^\circ = mg \]
Given: \[ m = 5g = 0.005\,kg,\quad g = 10\,m\,s^{-2} \] \[ mg = 0.005 \times 10 = 0.05\,N \]
So, \[ qE \times \frac{1}{2} = 0.05 \Rightarrow qE = 0.1\,N \]
Step 4: Apply horizontal equilibrium condition.
Tension balances horizontal electric component: \[ T = qE\cos 30^\circ \]
Step 5: Substitute values.
\[ T = 0.1 \times \frac{\sqrt{3}}{2} = 0.0866\,N \]
Step 6: Final conclusion.
Thus, the tension in the string is: \[ \boxed{8.66 \times 10^{-2}\,N} \]
The forces are: - Weight \(mg\) acting downward - Electric force \(qE\) acting at \(30^\circ\) above horizontal - Tension \(T\) acting horizontally (string is horizontal)
Step 2: Resolve electric force into components.
\[ (qE)_x = qE\cos 30^\circ,\quad (qE)_y = qE\sin 30^\circ \]
Step 3: Apply vertical equilibrium condition.
Since the string is horizontal, vertical forces balance only between weight and vertical electric component: \[ qE\sin 30^\circ = mg \]
Given: \[ m = 5g = 0.005\,kg,\quad g = 10\,m\,s^{-2} \] \[ mg = 0.005 \times 10 = 0.05\,N \]
So, \[ qE \times \frac{1}{2} = 0.05 \Rightarrow qE = 0.1\,N \]
Step 4: Apply horizontal equilibrium condition.
Tension balances horizontal electric component: \[ T = qE\cos 30^\circ \]
Step 5: Substitute values.
\[ T = 0.1 \times \frac{\sqrt{3}}{2} = 0.0866\,N \]
Step 6: Final conclusion.
Thus, the tension in the string is: \[ \boxed{8.66 \times 10^{-2}\,N} \]
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