Question #211037

The electric field intensity at a point situated 4 meters from a point charge is 200N/C. If the distance is reduced to 2 meters, the field intensity will be?


1
Expert's answer
2021-06-30T16:18:35-0400
E=kqr2E=2004222=800NCE=k\frac{q}{r^2}\\E'=200\frac{4^2}{2^2}=800\frac{N}{C}


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