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Induced Charges

What will be ...

Question

What will be the effect on capacitance of a charged capacitor when a metal plate is inserted between the plates?

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Solution

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$C a p a c i \tan c e o f a p a r a l l e l p l a t e c a p a c i t o r i s C_{1} = \frac{\in A}{d} C a p a c i \tan c e o f a p a r a l l e l p l a t e c a p a c i t o r w h e n f i l l e d w i t h a d i e l e c t r i c o f d i e l e c t r i c c o n s \tan t K i s C_{2} = \frac{\in A}{d - t + \frac{t}{K}} F o r a c o n d u c t o r d i e l e c t r i c c o n s \tan t i s K = \infty C_{2} = \frac{\in A}{d - t + \frac{t}{\infty}} = \frac{\in A}{d - t} C_{2} > C_{1} S o c a p a c i \tan c e i n c r e a s e s w i t h t h e i n t r o d u c t i o n o f c o n d u c t i n g s l a b . R e g a r d s$

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Similar questions

K = 5

50 μ F

200 μ C

A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:

A

d

\frac{d}{2}

A

C_{1}

C_{2}

\frac{C^{'}}{C}

\frac{d}{2}

2 μ F

10 V

K = 4

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