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.
For any further help, please write in the comments.
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