Asked by Jack
An electric field exerts a force of 3.00 E-4 N on a positive test charge of 7.20 E-4 C. The magnitude of the field at the location of the charge is?
Answers
Answered by
drwls
The field,. designated by E, satisfies the formula
F = Q*E
Therefore E = F/Q
The units will be Newtons per Coulomb, which is the same as Volts per meter
F = Q*E
Therefore E = F/Q
The units will be Newtons per Coulomb, which is the same as Volts per meter
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