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If the separation between the plates of ...

If the separation between the plates of a capacitor is 5 mm , then area of the plate of a 3 F parallel plate capacitor is

A

`4.259xx10^(9) m^(2)`

B

`1.964 xx10^(9) m^(2)`

C

`12.81 xx10^(9)m^(2)`

D

`1.694xx10^(9)m^(2)`

Text Solution

AI Generated Solution

To find the area of the plates of a parallel plate capacitor with a capacitance of 3 F and a separation of 5 mm, we can use the formula for the capacitance of a parallel plate capacitor: \[ C = \frac{\varepsilon_0 \cdot A}{d} \] Where: - \( C \) is the capacitance (in Farads), - \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{F/m} \)), - \( A \) is the area of the plates (in square meters), ...
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Knowledge Check

  • 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.

  • 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)`

  • The separation between the plates of a parallel plate capacitor is d and the area of each plate is A. iff a dielectric slab of thickness x and dielectric constant K is introduced between the plates, then the capacitance will be

    A
    `(epsi_(0)A)/(d-x(1-(1)/(K)))`
    B
    `(epsi_(0)A)/(d+x(1-(1)/(K)))`
    C
    `(epsi_(0)A)/(d+x(1+(1)/(K)))`
    D
    `(epsi_(0)A)/(d-x(1+(2)/(K)))`

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