Physics, asked by haione, 8 months ago

In a parallel plate capacitor, the capacitance increases by introducing a
dielectric in plates from
2uF to 20uF. The value of dielectric constant is:​

Answers

Answered by alam15hannan
1

Answer:

Initial capacitance is given as C=10μF

⇒C=

d

ε

0

A

=10μF

New arrangement acts as two capacitors C

1

and C

2

connected in parallel.

Thus capacitance of each part C

1

=

d

k

1

ε

0

A

1

and C

2

=

d

k

2

ε

0

A

2

Equivalent capacitance of parallel combination C

p

=C

1

+C

2

=(k

1

+k

2

)

d

ϵ

0

(A

1

+A

2

)

⟹ C

p

=(k

1

+k

2

)

2d

ϵ

0

A

(∵A

1

=A

2

=

2

A

)

Thus C

p

=(2+4)×

2

10

=30μF

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