Question:
If a spherical conductor of $10\,cm$ radius contains $5 \times 10^6$ electrons, then the electric field on its surface (in $NC^{-1}$) is
If a spherical conductor of $10\,cm$ radius contains $5 \times 10^6$ electrons, then the electric field on its surface (in $NC^{-1}$) is
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Always convert cm to meters before calculation.
Updated On: Apr 24, 2026
- $0.86$
- $0.36$
- $0.45$
- $1.44$
- $0.72$
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The Correct Option is
Solution and Explanation
Concept:
• Electric field: \[ E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \]
Step 1: Charge
\[ Q = ne = 5\times10^6 \times 1.6\times10^{-19} = 8\times10^{-13}C \]
Step 2: Radius
\[ r = 10cm = 0.1m \]
Step 3: Compute
\[ E = \frac{9\times10^9 \cdot 8\times10^{-13}}{(0.1)^2} \] \[ = \frac{72\times10^{-4}}{0.01} = 0.72 \] Final Conclusion:
Option (E)
• Electric field: \[ E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \]
Step 1: Charge
\[ Q = ne = 5\times10^6 \times 1.6\times10^{-19} = 8\times10^{-13}C \]
Step 2: Radius
\[ r = 10cm = 0.1m \]
Step 3: Compute
\[ E = \frac{9\times10^9 \cdot 8\times10^{-13}}{(0.1)^2} \] \[ = \frac{72\times10^{-4}}{0.01} = 0.72 \] Final Conclusion:
Option (E)
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