Question

a parallel plate condenser has two circular metal plates of radius 10 centimetre separated by certain distance the condenser is being charged with variable electric field at the rate of 5 into 10 ^ 13 volt per Mili second the displacement current is​

Answer 1

Answer:

Given: A parallel plate condenser consists of two circular plates each of radius 2cm separated by a distance of 0.1mm. A time varying potential difference of 5×10

13

v/s is applied across the plates of the condenser.

To find the displacement current

Solution:

Radius of each circular plate, r=2cm=0.02m

Distance between the plates, d=0.1mm=0.1×10

−3

m

the change in potential difference between the plates,

dt

dV

=5×10

13

vs

Permittivity of free space, ε

0

=8.85×10

−12

C

2

N

−1

m

−2

Hence area of the each plate, A=πr

2

=π(0.02)

2

=1.26×10

−3

m

2

......(i)

Capacitance between the two plates is given by the relation,

C=

d

ε

0

A

⟹C=

0.1×10

−3

8.85×10

−12

×1.26×10

−3

⟹C=11.15×10

−11

F

Now we know

displacement current,

I

d

=

dt

dq

where q is the charge on each plate and it is equal to q=CV

I

d

=C

dt

dV

By substituting the values, we get

I

d

=11.15×10

−11

×5×10

13

⟹I

d

=57.55×10

2

⟹I

d

=5.7×10

3

A

is the displacement current.