Question

A parallel plate capacitor has a capacity 80 x 10-6 when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of 30 v by wires the dielectric slab is then removed. Then the charge that passes now through the wire is

Answer 1

45.6×10−3C is the answer

Answer 2

Answer:

Charge that passes through the wire is

$45.6×10 ^{ - 3} c$

Explanation:

As data given in question:

C (Capacity of the capacitor) =

$80 \times 10 ^{ - 6}$

K (Dielectric Constant) = 20

V (Battery) = 30V

Charge passing through wire is given by,

Δq=ΔCV

where, Δq = charge passing through wire.

C = Capacity of capacitor

V = Battery

Using this formula let's derive a solution.

Δq=ΔCV

=(C′−C)V

= (k−1)CV

=(20−1)(80×10^{ - 6} )(30)

=45.6×10 ^{ - 3}

Conclusion:

Thus, the charge that passes through wire is

$45.6×10 ^{ - 3} c$