Question:
A particle of charge \(7 \, \mu C\) is moved away from an infinite non-conducting sheet of surface charge density \(5.0 \, \mu C\, m^{-2}\) to a point at a distance of \(5.0 \, cm\). The work done by the field due to the sheet is \(\left(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, SI \, units\right)\)
A particle of charge \(7 \, \mu C\) is moved away from an infinite non-conducting sheet of surface charge density \(5.0 \, \mu C\, m^{-2}\) to a point at a distance of \(5.0 \, cm\). The work done by the field due to the sheet is \(\left(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, SI \, units\right)\)
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Electric field of an infinite sheet is constant, so work depends only on displacement, not path.
Updated On: Jul 18, 2026
- 79 mJ
- 158 mJ
- 49 mJ
- 99 mJ
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The Correct Option is D
Solution and Explanation
Step 1: Electric field due to infinite non-conducting sheet.
For an infinite non-conducting sheet with surface charge density \(\sigma\), the electric field is constant and given by: \[ E = \frac{\sigma}{2\varepsilon_0} \] This field is uniform and does not depend on distance.
Step 2: Understanding work done by electric field.
Work done by electric field when a charge moves in uniform field is: \[ W = q E d \] where \(d\) is displacement along field direction.
Step 3: Substituting given values.
\[ q = 7 \times 10^{-6} C,\quad \sigma = 5 \times 10^{-6} C/m^2,\quad d = 5 \times 10^{-2} m \] Electric field: \[ E = \frac{5 \times 10^{-6}}{2\varepsilon_0} \]
Step 4: Using \(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\).
We use: \[ \frac{1}{\varepsilon_0} = 4\pi \times 9 \times 10^9 \] So: \[ E = \frac{5 \times 10^{-6}}{2} \cdot (4\pi \cdot 9 \times 10^9) \]
Step 5: Compute work done.
\[ W = qEd \] Substituting and simplifying gives: \[ W = 9.9 \times 10^{-2} \, J \]
Step 6: Final result.
\[ W = 99 \, mJ \] \[ \boxed{99 \, mJ} \]
For an infinite non-conducting sheet with surface charge density \(\sigma\), the electric field is constant and given by: \[ E = \frac{\sigma}{2\varepsilon_0} \] This field is uniform and does not depend on distance.
Step 2: Understanding work done by electric field.
Work done by electric field when a charge moves in uniform field is: \[ W = q E d \] where \(d\) is displacement along field direction.
Step 3: Substituting given values.
\[ q = 7 \times 10^{-6} C,\quad \sigma = 5 \times 10^{-6} C/m^2,\quad d = 5 \times 10^{-2} m \] Electric field: \[ E = \frac{5 \times 10^{-6}}{2\varepsilon_0} \]
Step 4: Using \(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\).
We use: \[ \frac{1}{\varepsilon_0} = 4\pi \times 9 \times 10^9 \] So: \[ E = \frac{5 \times 10^{-6}}{2} \cdot (4\pi \cdot 9 \times 10^9) \]
Step 5: Compute work done.
\[ W = qEd \] Substituting and simplifying gives: \[ W = 9.9 \times 10^{-2} \, J \]
Step 6: Final result.
\[ W = 99 \, mJ \] \[ \boxed{99 \, mJ} \]
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