A 23900 N/C electric field points down. What is the value of a charge placed in the field if it experiences a force of 7520N up?
A)3.18C
B)0.315C
C)-3.18C
D)-0.315C
Answers
Answered by
1
As we know F = qE
we have given : E = 23900N/C , F = 7520 N
7520 = 23900 q
q = 7520/ 23900
q = 0.315 C
Similar questions