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. 79 mJ
• B. 158 mJ
• C. 49 mJ
• D. 99 mJ

Hint:

Electric field of an infinite sheet is constant, so work depends only on displacement, not path.

Answer 1

The Correct Option is D

Step 1: Recall the field of an infinite charged sheet.
An infinite non-conducting sheet with surface charge density $\sigma$ gives a uniform field on either side, $E = \frac{\sigma}{2\varepsilon_0}$, that does not weaken with distance.

Step 2: Work out $\varepsilon_0$ as a decimal from the given constant, instead of carrying $\frac{1}{4\pi\varepsilon_0}$ through the algebra symbolically.
\[ \varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} \approx 8.84 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2} \]
Step 3: Compute the field numerically.
\[ E = \frac{5.0 \times 10^{-6}}{2 \times 8.84 \times 10^{-12}} = \frac{5.0 \times 10^{-6}}{1.77 \times 10^{-11}} \approx 2.83 \times 10^{5}\ \text{N C}^{-1} \]
Step 4: Since the field is uniform, work done equals force times distance directly.
\[ W = qEd = (7 \times 10^{-6}) \times (2.83 \times 10^{5}) \times (5.0 \times 10^{-2}) \]
Step 5: Work through the multiplication.
\[ 7 \times 10^{-6} \times 2.83 \times 10^{5} \approx 1.98,\qquad 1.98 \times 5.0 \times 10^{-2} \approx 9.9 \times 10^{-2}\ \text{J} \]
Step 6: Convert to millijoules.
\[ 9.9 \times 10^{-2}\ \text{J} = 99\ \text{mJ} \]
Final Answer:
\[ \boxed{99\ \text{mJ}} \]