Practicing Success
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$\text{Electric Field due to rod is at its surface } E = \frac{\sigma}{2\epsilon_0} = \frac{50\times 10^{-6}}{2\times 8.85\times 10^{-12}} = 2.82 \times 10^6 N/C$ $ F = qE = \sigma A E = 50\times 10^{-6} \times 10^{-2} \times 2.82 \times 10^6 = 1.4N$ |
