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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?

  • Physics -

    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

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