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A positive charge of 1 nC is placed at (0, 0, 0.2) where all dimensions are in metres. Consider the x-y plane to be a conducting ground plane. Take ε0 = 8.85×10⁻¹² F/m. The z component of the E field at (0, 0, 0.1) is closest to

1. 899.18 V/m
2. -899.18 V/m
3. 999.09 V/m
4. -999.09 V/m

Correct answer: -999.09 V/m

Solution

By image theory, place -1 nC at z=-0.2. At (0,0,0.1) the real charge (0.1 m away) gives E_z=-899.18 V/m and the image charge (0.3 m away) gives E_z=-99.91 V/m, both in the -z direction. The total is about -999.09 V/m.

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