Answer :


Given,


Charge on plate 1, Q1 = +2.0 × 10–8C


Charge on plate 2, Q2 = –1.0 × 10–8C


Capacitance of the capacitor, C= 1.2 × 10–3 μF= 1.2 × 10–9 F


Formula used


We know that for a capacitor with net charge, Q and capacitance, C, the Potential difference deceloped in between the plates, V is,



The charges on the inner plates of the capacitor with plates having charges Q1 and Q2 is,



Note: Charges on the outer plates of the capacitor with plates having charges Q1 and Q2 is,



In the given example, the plates has individual charges Q1 and Q2. But when they placed as a capacitor, their charges re-arrange and equal and opposite charges will be distributed in each plates. The total net charge, Qnet on the inner sides of each plates will be



Substituting the values,




Hence the inner side of each plates will have a charge of ±1.5 × 10–8C.


Hence from eqn.1, Potential difference, V is



Or,



Hence the potential difference developed between the plates is 12.5V


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