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A parallel plate capacitor of area A and...

A parallel plate capacitor of area A and separation d is charged to potential difference V and removed from the changing source. A dielectric slab of constant K =2, thickness d and are area `(A)/(2)` is inserted, as shown in the figure. Let `sigma_(1)` be free charge density at the conductor-dielectric surface and `sigma_(2)` be the charge density at the conductor-vacuum surface

A

The electric field have same value inside the dielectric as in the free space between the plates

B

The ratio `(sigma_(1))/(sigma_(2))` is equal to `(2)/(1)`

C

The new capcitance is `(3 in_(0) A)/(2d)`

D

The new potential difference is `(2)/(3) V`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Potential for each plate remain same over whole area. If potential difference between them is, say V' then V' = Ed
i.e. E is also same inside the plates to keep E same, free charge density is changed i.e charge redistributes itself
To find new capacitance two capacitors can be taken as connected in parallel. Then
`-(K.in_(0)A//2)/(d) +(in_(0)A//2)/(d)=(3 in_(0)A)/(2d)`
By `Q = CV`, as Q remains unchanged V is changed to `(2)/(3) V`.
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