Question

In a circular parallel plate capacitor radius of each plate is 5cm and they are separated by a distance of 2mm calculate the capacitance and the energy stored, when it is charged by connecting the battery of 200v

Answer 1

Capacitance of the capacitor,

C = (Fsilen knot × area) ÷ distance between the plates

C = (8.854×10^-12×3.14×0.05×0.05) ÷ 0.002

C = 34.76pF

Now, energy

E = ½CV²

E = ½×34.76×10^-12×200×200

E = 34.76 × 2 × 10^-8

E = 6.95×10^-7 J

Answer 2

The capacitance and the energy stored in the capacitor is $3.47\times 10^{-11}\ F$ and $6.94\times 10^{-7}\ J$.

Explanation:

It is given that,

Radius of each plate, r = 5 cm = 0.05 m

Separation between plates, d = 2 mm = 0.002 m

If C is the capacitance of the parallel plate capacitor. We know that the formula of capacitance of a capacitor is given by :

$C=\dfrac{A\epsilon_o}{d}$

$C=\dfrac{\pi r^2\epsilon_o}{d}$

$C=\dfrac{\pi (0.05)^2\times 8.85\times 10^{-12}}{0.002}$

$C=3.47\times 10^{-11}\ F$

Voltage of the battery, V = 200 V

The energy stored in the capacitor is given by :

$E=\dfrac{1}{2}CV^2$

$E=\dfrac{1}{2}\times 3.47\times 10^{-11}\times (200)^2$

$E=6.94\times 10^{-7}\ J$

So, the capacitance and the energy stored in the capacitor is $3.47\times 10^{-11}\ F$ and $6.94\times 10^{-7}\ J$. Hence, this is the required solution.

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