# A capacitor havin

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 :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

A capacitor is maPhysics - Exemplar

A parallel plate Physics - Exemplar

The battery remaiPhysics - Exemplar

A parallel-plate HC Verma - Concepts of Physics Part 2

How many time conHC Verma - Concepts of Physics Part 2

The plates of a cHC Verma - Concepts of Physics Part 2

A capacitor of caHC Verma - Concepts of Physics Part 2