8. (10 points) A parallel plate capacitor has plate area A, the plates are separated by...

SOLUTION ==>

For a parallel plate capacitor,

Voltage between the plates = V = Q/C

Q => Charge on the capacitor

C => Capacitance of the capacitor

Since, the direction of this electric field goes from positive to negative plate of the capacitor, therefore, for a positive test charge q, the force exerted by the field will be such that the test charge q moves towards the negative plate. Hence, the external work done by us, which is moving the test charge from negative plate to positive plate, will do a positive work. Let this external force applied by us = F

Here, the assumption of q being a test charge was used.

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For a parallel plate capacitor, Voltage between the plates - v = Q/ c Q = Charge On the capacitor C = Capacitance of the capacitor v = Q/C C = elementof/ A/s Therefore, Elective field between plates == e = V/ s = Q/ Cs E = Q/ element_0 A Therefore, force exerted by electric field on this test charge = q E = Qq/elementof A Since, the direction of this electric field goes from positive to negative plate of the capacitor, therefore, for a positive test charge q, the force exerted by the field will be such that the test charge q moves towards the negative plate. Hence, the external work done by us, which is moving the test charge from negative plate to positive plate, will do a positive work. Let this external force applied by us = F Therefore, F = qE = Qq/ elementof A Required work done = F. s = Qqs/ element_0 A