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A parallel plate capacitor filled with a dielectric of relative permittivity 5 between its plates is charged to acquire an energy E and isolated. If the dielectric is replaced by another of relative permittiviey 2, its energy becomes
- a)E
- b)0.4 E
- c)2.5 E
- d)6.25 E
Correct answer is option 'C'. Can you explain this answer?
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A parallel plate capacitor filled with a dielectric of relative permit...
E ∝ 1/k
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A parallel plate capacitor filled with a dielectric of relative permit...
Given data:
Relative permittivity of dielectric between plates initially (ε1) = 5
Relative permittivity of dielectric between plates after replacement (ε2) = 2
Initial energy stored in capacitor (E1) = E
Explanation:
1. Relationship between energy stored and capacitance:
The energy stored in a capacitor is given by the formula:
E = 1/2 * CV^2
Where,
E = energy stored
C = capacitance
V = voltage across the capacitor
2. Effect of dielectric on capacitance:
The presence of a dielectric material between the plates of a capacitor increases the capacitance. The relationship between the capacitance with and without the dielectric is given by:
C = ε0 * εr * A / d
Where,
C = capacitance
ε0 = permittivity of free space
εr = relative permittivity of the dielectric
A = area of the plates
d = distance between the plates
3. Calculation of energy with new dielectric:
Initially, the energy stored in the capacitor is given by:
E1 = 1/2 * C1 * V^2
After replacing the dielectric with εr = 2, the new capacitance becomes:
C2 = ε0 * 2 * A / d = 2 * C1
The energy stored in the capacitor with the new dielectric is given by:
E2 = 1/2 * C2 * V^2 = 1/2 * 2 * C1 * V^2 = 2 * (1/2 * C1 * V^2) = 2 * E1 = 2E
Conclusion:
Therefore, the energy stored in the capacitor after replacing the dielectric with relative permittivity 2 becomes 2 times the initial energy stored in the capacitor. Hence, the correct answer is option C: 2.5 E.
Relative permittivity of dielectric between plates initially (ε1) = 5
Relative permittivity of dielectric between plates after replacement (ε2) = 2
Initial energy stored in capacitor (E1) = E
Explanation:
1. Relationship between energy stored and capacitance:
The energy stored in a capacitor is given by the formula:
E = 1/2 * CV^2
Where,
E = energy stored
C = capacitance
V = voltage across the capacitor
2. Effect of dielectric on capacitance:
The presence of a dielectric material between the plates of a capacitor increases the capacitance. The relationship between the capacitance with and without the dielectric is given by:
C = ε0 * εr * A / d
Where,
C = capacitance
ε0 = permittivity of free space
εr = relative permittivity of the dielectric
A = area of the plates
d = distance between the plates
3. Calculation of energy with new dielectric:
Initially, the energy stored in the capacitor is given by:
E1 = 1/2 * C1 * V^2
After replacing the dielectric with εr = 2, the new capacitance becomes:
C2 = ε0 * 2 * A / d = 2 * C1
The energy stored in the capacitor with the new dielectric is given by:
E2 = 1/2 * C2 * V^2 = 1/2 * 2 * C1 * V^2 = 2 * (1/2 * C1 * V^2) = 2 * E1 = 2E
Conclusion:
Therefore, the energy stored in the capacitor after replacing the dielectric with relative permittivity 2 becomes 2 times the initial energy stored in the capacitor. Hence, the correct answer is option C: 2.5 E.
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A parallel plate capacitor filled with a dielectric of relative permittivity 5 between its plates is charged to acquire an energy E and isolated. If the dielectric is replaced by another of relative permittiviey 2, its energy becomesa)Eb)0.4 Ec)2.5 Ed)6.25 ECorrect answer is option 'C'. Can you explain this answer?
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