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, q_{i} is the induced charge, q is the initial charge and k is the dielectric constant of the material inserted.

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