Answer :

Given,


Charge given to the upper plate, plate P, is 1.0 μC.


Formula used


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,



a)


By giving a charge of 1.0 μC to plate P, it will get distributed on either side of the plate as +0.5 μC. The other plates get induced with this charge as shown in figure.





To find the charge on the plate Q, eqn.2 shall be used. So we get,



Where Q1 is the charge on one plate P= 1.0 μC


And Q2 is the charge on plate Q = 0C


Hence by substituting in the above equation, we get,



So the upper face of plate Q will get a charge of -0.5 μC and this will induce a charge of +0.5 μC on the bottom side of plate Q.


Hence the potential difference in capacitor P-Q, by eqn.1 is



Where Q= 0.5 μC


And C= 50×10-3μF


Hence,



So the potential difference in between the upper and middle plates is 10V


b)


From the above condition, the upper face of plate Q will get a charge of -0.5 μC and this will induce a charge of +0.5 μC on the bottom side of plate Q.


We know that, the capacitor Q-R is made of the bottom surface of plate Q and the upper side of plate R. As the bottom surface of plate Q already has a charge of +0.5 μC, it will induce -0.5 μC charge on the upper face of plate R As shown in figure).


Hence the potential difference in between the lower and middle plates can be calculated from the eqn.1, as



Where Q is the charge in each plates=±0.5 μC


And C= 50×10-3μF


Hence,



So the potential difference in between the middle and lower plates is 10V


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