Posted by Jack on .
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?
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