Question #211049

If the potential difference between two points 2cm apart in an electric field is 20V, the electric field intensity between these points will be around


1
Expert's answer
2021-07-05T08:29:29-0400

The electric field intensity is given by

E=Vd=20V0.02m=1000V/mE=\frac{V}{d}=\frac{20\:\rm V}{0.02\:\rm m}=1000\:\rm V/m


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