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A capacitor of capacitance C is charged ...

A capacitor of capacitance `C` is charged to a potential `V`. The flux of the electric field through a closed surface enclosing the capacitor is

A

`(CV)/(epsilon)`

B

`(2 CV)/(epsilon_0)`

C

`(CV)/(2espsilon_0)`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
D
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Knowledge Check

  • A capacitor of capacitance C_(1) is kept charged to a potential V. If this capacitor is then connected in parallel to an uncharged capacitor of capacitance C_(2) then fin the final potential difference across the combination.

    A
    `(C_(1)V)/(C_(1)+C_(2))`
    B
    `(C_(2)V)/(C_(1)+C_(2))`
    C
    `1+(C_(2))/(C_(1))`
    D
    `1-(C_(2))/(C_(1))`

  • In the circuit shown in figure-3.345 the capacitor of capacitance C is charged to a potential difference V. The current in the circuit just after the closing of switch S is :

    A
    `(V)/(3R)`
    B
    `(3V)/(R)`
    C
    `(V)/(2R)`
    D
    zero

  • A capacitor of capacitance C is charged to a potential difference V_(0) . The charged battery is disconnected and the capacitor is connected to a capacitor of unknown capacitance C_(x) . The potential difference across the combination is V. The value of C_(x) should be

    A
    `(C(V_(0) - V))/(V)`
    B
    `(C(V - V_(0)))/(V)`
    C
    `(CV)/(V_(0))`
    D
    `(CV_(0))/(V)`

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