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.
Rate this question :


A capacitor is ma
Physics - ExemplarA parallel plate
Physics - ExemplarThe battery remai
Physics - ExemplarA parallel-plate
HC Verma - Concepts of Physics Part 2How many time con
HC Verma - Concepts of Physics Part 2The plates of a c
HC Verma - Concepts of Physics Part 2A capacitor of ca
HC Verma - Concepts of Physics Part 2