Question

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 5x1013 V/s is applied
across the plates of the condenser. The
displacement current is
1) 5.50 A
2) 5.56x102 A
3) 5.56x103 A
4) 2.28 x 104 A​

Answer 1

Given:

• Distance between the parallel plates = 0.1 mm = 0.0001 m
• Radius of each plate = 2 cm = 0.02 m
• Change in potential difference with time = 5 x 10¹³ V/ s

To find:

• Displacement current

Method:

Step I: Find capacitance.

The capacitance is given by,

$\blacksquare$ C = E• A / d

here,

• C = capacitance
• E• = permitivitty = 8.8 x 10⁻¹² F/m
• A = area
• d = distance

⇒ C = E• x π r ² / d

⇒ C = 8.8 x 10⁻¹² x 3.14 x 4 x 10⁻⁴ / 1 x 10⁻⁴

⇒ C = 110.52 x 10⁻¹² F

Step II: Find displacement current

We know that,

$\blacksquare$ I = dq/dt

And,

• q = CV

Hence,

⇒ I = d(CV)/ dt

⇒ I = C dV/ dt

Now let's substitute the values,

⇒ I = 110.52 x 10⁻¹² x 5 x 10¹³

⇒ I = 552.6 x 10

⇒ I = 5526 A

⇒ I = 5.526 x 10³ A

Hence,

The answer is option 3.