A capacitor,consisting two metal plates each of area 50cm2 and spaced 0.2mm apart in air ,is connected across a 120V supply.Calculate (1)the energy stored (2)the ekectric flux density and the potential gradient..

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Physics

To calculate the energy stored in a capacitor, you can use the formula:

Energy (E) = 1/2 * C * V^2

where C is the capacitance and V is the voltage.

1. Capacitance (C):
The capacitance of a parallel plate capacitor can be calculated using the formula:

C = ε₀ * (A/d)

where ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 F/m), A is the area of one plate, and d is the separation between the plates.

First, we need to convert the given area from cm² to m²:
Area (A) = 50 cm² = 50 * (0.01 m)^2 = 0.05 m²

Using the given separation of 0.2 mm, convert it to meters:
Separation (d) = 0.2 mm = 0.2 * (0.001 m) = 0.0002 m

Now, we can calculate the capacitance:
C = (8.85 x 10^-12 F/m) * (0.05 m²/0.0002 m) = 2.21 x 10^-8 F

2. Energy (E):
Given that the capacitor is connected to a 120V supply, we can now calculate the energy stored in the capacitor:
E = 1/2 * (2.21 x 10^-8 F) * (120 V)^2 = 0.159 J (joules)

To calculate the electric flux density (D) and the potential gradient (E field), we can use the formulas:

D = Q/A
E = V/d

where Q is the charge, A is the area, V is the voltage, and d is the separation between the plates.

The charge (Q) stored in a capacitor can be calculated using the formula:
Q = C * V
where C is the capacitance and V is the voltage.

1. Electric Flux Density (D):
D = Q/A

Calculate the charge:
Q = (2.21 x 10^-8 F) * (120 V) = 2.652 x 10^-6 C (coulombs)

D = (2.652 x 10^-6 C) / (0.05 m²) = 5.304 x 10^-5 C/m²

2. Potential Gradient (E field):
E = V/d

E = (120 V) / (0.0002 m) = 6 x 10^5 V/m (volts per meter)

So, the answers are:
(1) The energy stored in the capacitor is 0.159 J (joules).
(2) The electric flux density is 5.304 x 10^-5 C/m² and the potential gradient (E field) is 6 x 10^5 V/m.