Question

An air capacitor of capacity c=10uf is connected to a constant voltage battery of 10v. now the space between the plates is filled with a liquid of dielctric constant5.calculate additional charge which flows from battery to capacitor

Answer 1

Hey Dear,

◆ Answer -

∆Q = 0.4 C

● Explaination -

# Given -

C = 10 uF

V = 10 V

k = 5

# Solution -

Initially, charge flowing through the battery of air capacitor is given by -

Q = C.V

Q = 10 × 10

Q = 100 uC

When capaitor plates are filled with dielectric liquid.

C' = k.C

C' = 5 × 10

C' = 50 uF

Finally, charge flowing through the battery of liquid capacitor is given by -

Q' = C'.V

Q' = 50 × 10

Q' = 500 uC

Increase in charge is calculated as -

∆Q = Q' - Q

∆Q = 500 - 100

∆Q = 400 uC

∆Q = 0.4 C

Therefore, additional charge flowing from battery to capacitor is 0.4 C.

Thanks dear...

Answer 2

Answer:

Explanation:

Original charge=10*10=100uC=Qo

Dielectric constant=5=k

New charge=Qo(1-(1/k)) (Formula)

=100(1-(1/5)=80uC

So, additional charge=100-80=20uC, hope you understood.