A capacitor having a capacitance of 100 μF is charged to a potential difference 50V.
a) What is the magnitude of the charge on each plate?
b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates.
c) What charge would have produced this potential difference in absence of the dielectric slab.
d) Find the charge induced at a surface of the dielectric slab.
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.
