Question:
An infinite non-conducting sheet has a surface charge density of
\[
7\times 10^{-7}\ \text{C m}^{-2}
\]
on one side. The distance between equipotential surfaces whose potentials differ by \(19.8\ \text{V}\), will be
\[
\left(\frac{1}{4\pi\varepsilon_0}=9\times 10^9\ \text{SI units}\right)
\]
An infinite non-conducting sheet has a surface charge density of
\[
7\times 10^{-7}\ \text{C m}^{-2}
\]
on one side. The distance between equipotential surfaces whose potentials differ by \(19.8\ \text{V}\), will be
\[ \left(\frac{1}{4\pi\varepsilon_0}=9\times 10^9\ \text{SI units}\right) \]
\[ \left(\frac{1}{4\pi\varepsilon_0}=9\times 10^9\ \text{SI units}\right) \]
Show Hint
For an infinite non-conducting sheet:
\[
E=\frac{\sigma}{2\varepsilon_0}
\]
and for a uniform electric field:
\[
V=Ed
\]
Use these two formulas together to find separation between equipotential surfaces.
Updated On: Jun 25, 2026
- \(2.0\ \text{mm}\)
- \(0.25\ \text{mm}\)
- \(1.0\ \text{mm}\)
- \(0.5\ \text{mm}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the electric field due to an infinite non-conducting sheet.
The electric field due to an infinite non-conducting sheet is \[ E=\frac{\sigma}{2\varepsilon_0} \] Given: \[ \sigma=7\times 10^{-7}\ \text{C m}^{-2} \] Also, \[ \frac{1}{4\pi\varepsilon_0}=9\times 10^9 \] Using \[ \frac{1}{\varepsilon_0}=4\pi\times 9\times 10^9 \] we get \[ E=\frac{\sigma}{2\varepsilon_0} \] \[ =2\pi \left(9\times 10^9\right)\sigma \] Substituting \(\sigma\), \[ E=2\pi(9\times 10^9)(7\times 10^{-7}) \] \[ E=18\pi\times 7\times 10^2 \] \[ E=126\pi\times 10^2 \] \[ E\approx 3.96\times 10^4\ \text{V/m} \]
Step 2: Use relation between potential difference and distance.
For uniform electric field, \[ V=Ed \] Hence, \[ d=\frac{V}{E} \] Given potential difference: \[ V=19.8\ \text{V} \] Therefore, \[ d=\frac{19.8}{3.96\times 10^4} \] \[ d=5\times 10^{-4}\ \text{m} \]
Step 3: Convert into millimeters.
Since \[ 1\ \text{mm}=10^{-3}\ \text{m}, \] \[ d=0.5\ \text{mm} \]
Step 4: Final conclusion.
Hence, the required distance is \[ \boxed{0.5\ \text{mm}} \]
The electric field due to an infinite non-conducting sheet is \[ E=\frac{\sigma}{2\varepsilon_0} \] Given: \[ \sigma=7\times 10^{-7}\ \text{C m}^{-2} \] Also, \[ \frac{1}{4\pi\varepsilon_0}=9\times 10^9 \] Using \[ \frac{1}{\varepsilon_0}=4\pi\times 9\times 10^9 \] we get \[ E=\frac{\sigma}{2\varepsilon_0} \] \[ =2\pi \left(9\times 10^9\right)\sigma \] Substituting \(\sigma\), \[ E=2\pi(9\times 10^9)(7\times 10^{-7}) \] \[ E=18\pi\times 7\times 10^2 \] \[ E=126\pi\times 10^2 \] \[ E\approx 3.96\times 10^4\ \text{V/m} \]
Step 2: Use relation between potential difference and distance.
For uniform electric field, \[ V=Ed \] Hence, \[ d=\frac{V}{E} \] Given potential difference: \[ V=19.8\ \text{V} \] Therefore, \[ d=\frac{19.8}{3.96\times 10^4} \] \[ d=5\times 10^{-4}\ \text{m} \]
Step 3: Convert into millimeters.
Since \[ 1\ \text{mm}=10^{-3}\ \text{m}, \] \[ d=0.5\ \text{mm} \]
Step 4: Final conclusion.
Hence, the required distance is \[ \boxed{0.5\ \text{mm}} \]
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