A parallel plate capacitor has circular plates of 10 cm radius separated by an air-gap of 1 mm. It is charged by connecting the plates to a 100 volt battery. Then the change in energy stored in the capacitor when the plates are moved to a distance of 1 cm and the plates are maintained in connection with the batteryis:a)loss of 12.5 ergsb)loss of 125 ergsc)gain of 125 ergsd)gain of 12.5 ergsCorrect answer is option 'A'. Can you explain this answer?

Question

Answer

Solution:

Given, radius of circular plates, r = 10 cm = 0.1 m
Distance between the plates, d = 1 mm = 0.001 m
Potential difference, V = 100 V

1. Calculation of capacitance of the capacitor:

The capacitance of a parallel plate capacitor is given by the formula:

C = ε0A/d

where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

Here, the plates are circular in shape, so the area of each plate is given by:

A = πr^2

Substituting the given values, we get:

A = π(0.1)^2 = 0.0314 m^2

Using the formula for capacitance, we get:

C = ε0A/d = (8.85 × 10^-12)(0.0314)/(0.001) = 2.77 × 10^-10 F

2. Calculation of initial energy stored in the capacitor:

The energy stored in a capacitor is given by the formula:

U = (1/2)CV^2

Substituting the given values, we get:

U = (1/2)(2.77 × 10^-10)(100)^2 = 1.39 × 10^-6 J

3. Calculation of final energy stored in the capacitor:

When the plates are moved to a distance of 1 cm, the new distance between the plates is:

d' = 0.01 m

The capacitance of the capacitor in this case is given by:

C' = ε0A/d' = (8.85 × 10^-12)(0.0314)/(0.01) = 2.78 × 10^-11 F

The potential difference between the plates remains the same at 100 V.

Using the formula for energy stored in a capacitor, we get:

U' = (1/2)C'V^2 = (1/2)(2.78 × 10^-11)(100)^2 = 1.39 × 10^-6 J

4. Calculation of change in energy stored in the capacitor:

The change in energy stored in the capacitor is given by:

ΔU = U' - U

Substituting the calculated values, we get:

ΔU = (1.39 × 10^-6) - (1.39 × 10^-6) = 0 J

The change in energy stored in the capacitor is zero, which means there is no gain or loss of energy. However, as the plates are maintained in connection with the battery, some energy is lost in the form of heat due to the resistance of the wires and the battery. This loss of energy is given by:

ΔU = (1/2)CV^2 - (1/2)C'V^2

Substituting the calculated values, we get:

ΔU = (1/2)(2.77 × 10^-10)(100)^2 - (1/2)(2.78 × 10^-11)(100)^2 = 1.39 × 10^-7 J

Converting this to ergs, we get:

ΔU = 1.39 × 10^-7 × 10^7 =

Answer

C1=epsilonnot×A/d
C1= epsilonnot×π×10^-2/10^-3 = 10π×epsilonnot
U1=1/2×C1×V^2= 1/2×10π×epsilonnot×100^2=5×10^4π×epsilonnot
in the same way calculate C2 and U2
C2=epsilonnot×π×10^-2/10^-2= π×epsilonnot
U2=1/2×C2×V^2=1/2×π×epsilonnot×100×100=0.5×10^4π×epsilonnot
then take difference between U1&U2 we get 125.05×10^-8J
1J =10^7erg so 125.05×10^-8J is equal to 12.5 erg.. so this is the loss of energy because U2 is less than U1.

Similar Class 12 Doubts

A 2mf capacitor c1 is charged to a voltage of 100 v and a 4mf capacitor c2 is charged to a voltage of 50 v the capacitor are then connected in parallel.what is the loss of energy in parallel connection?

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