Answer :

Given –


Capacitance, C = 100 μF


Potential difference, V = 50V.


a) We know the magnitude of the charge on each plate is given by


Charge stored on the capacitor, q = c × v


where c is the capacitance and v is the potential difference


q = 100 μF×50


=5 mC


b)Now, the charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted.


The new potential difference between the plates will be –





c)


Now, the charge on the capacitance can be calculated as:


Charge, q= Capacitance, C × Potential difference, V


Putting the value in the above formula, we get


Q= 20 × 100 × 10-6 =2 mC


d)The charge induced at a surface of the dielectric slab –



Where, qi is the induced charge, q is the initial charge and k is the dielectric constant of the material inserted.




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