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The separation between the plates of a c...

The separation between the plates of a charged parallel-plate capacitor is increased. The force between the plates

A

increases

B

decreases

C

remains same

D

first increases then decreases

Text Solution

Verified by Experts

The correct Answer is:
C
`F=qE=q(sigma/(2epsilon_0))=q(q/(2Aepsilon_0))q` will not change
`:.` F=constant
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Knowledge Check

  • Force of attraction between the plates of a parallel plate capacitor is

    A
    `(q^(2))/(2 epsilon_(0)AK)`
    B
    `(q^(2))/(epsilon_(0)AK)`
    C
    `(q)/(2 epsilon_(0)A)`
    D
    `(q^(2))/(2 epsilon_(0)A^(2)K)`

  • As the distance between the plates of a parallel plate capacitor decreased

    A
    chances of electrical break down will increases if potential difference between the plates is kept constant.
    B
    chance of electrical break down will decreases if potential difference between the plates is kept constant.
    C
    chance of electric break down will increases if charge on the plates is kept constant
    D
    chance of electrical break down will decrease if charge on the plate is kept constant.

  • STATEMENT-1 : If distance between plates of charged and isolated parallel plate capacitor increases then force between plates decreases. and STATEMENT-2 : Force between two point charges is inversely proportional to the square of distance of separation.

    A
    Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
    B
    Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1
    C
    Statement-1 is True, Statement-2 is False
    D
    Statement-1 is False , Statement-2 is True

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