Electrostatics

Question: What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,newton/coulomb$ $[1/4\pi {\varepsilon _0} = 9 \times {10^9}N{m^2}/{C^2}]$

1) $2 \times {10^{ - 11}}coulomb$

2) $3 \times {10^{ - 11}}coulomb$

3) $5 \times {10^{ - 11}}coulomb$

4) $9 \times {10^{ - 11}}coulomb$

Solution:

Explanation: Explanation

electric-field-potential

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