Question

An air capacitor consists of a parallel square plates of 50 cm side and is charged to a potential difference of 250 V, when plates are 1mm apart. Find the work done in separating the plates from 1 to 3 mm. Assume perfect insulation.​

Answer 1

Answer:

We have C=Q/V

or Q=CV ,

for parallel plate capacitor , C=ε

0

KA/d ,

where A= area of a plate ,

d= separation between plates ,

and E=V/d or V=Ed

therefore Q=ε

0

KA/d×Ed ,

or Q=ε

0

KA×E

now charge on a plate Q will be maximum when electric field E is maximum ,

maximum value of E=3.0×10

4

V/cm=3.0×10

6

V/m ,

area of a plate A=5.0

2

=25cm

2

=25×10

−4

m

2

,

therefore maximum charge ,

Q=8.854×10

−12

×1×25×10

−4

×3.0×10

6

or Q=6.6×10

−8

C

y, Hope it helps u keep smiling