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What is the electric field in N/C) at a point where the force on a -2.1 x 10-5 C charge is (4.31 -6.89) x 10-6 N?
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What is the electric field in N/C) at a point where the force on a -2.1 x 10-5 C charge is (4.31 -6.89) x 10-6 N?

User Nikerboker
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1 Answer

7 votes

Answer: 0.123 N/C

Explanation: In order to solve this question we have to use the electric Force on a particle produced by an electric field which is given by:

F=q*E

so E=F/q= -2,58* 10^-6/-2.1*10^-5= 0.123 N/C

User Canned Man
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