A parallel-plate capacitor is charged by a battery and then the battery is removed and a dielectric of constant K is used to fill the gap between the plates. Inserting the dielectric changes the energy stored by a factor of
A)2.
B)1/K.
C)1 (no change).
D)K.
E)K-1.
Q could not change, no flow of current
but the initial capacitance Ci may change
Ui = .5 Q^2 / Ci
so what happens to C ?
K = C/Ci
so now C = K Ci
Uf = .5 Q^2 / (K Ci)
so
Uf/Ui = .5 Q^2 / (K Ci) /.5 Q^2 / (Ci)
= 1/K or B)
The energy stored in a capacitor is given by the formula:
U = (1/2) * C * V^2
Where U is the energy stored, C is the capacitance, and V is the voltage across the capacitor.
When a dielectric material of constant K is inserted between the plates of a capacitor, the capacitance increases by a factor of K. Therefore, the new capacitance C' is given by:
C' = K * C
Since the voltage across the capacitor does not change when the dielectric is inserted, the energy stored can be expressed as:
U' = (1/2) * C' * V^2
Substituting the value for C', we get:
U' = (1/2) * K * C * V^2
Comparing U' to the initial energy U, we can see that:
U' = K * U
Therefore, inserting the dielectric changes the energy stored by a factor of K.
The answer is D) K.
To find the change in energy stored in a parallel-plate capacitor when a dielectric of constant K is inserted between the plates, we need to understand the relationship between the energy stored in the capacitor and the dielectric constant.
The capacitance of a parallel-plate capacitor can be calculated using the formula:
C = ε₀A/d,
where C is the capacitance, ε₀ is the vacuum permittivity, A is the area of the plates, and d is the distance between the plates.
The energy stored in a capacitor is given by the formula:
U = (1/2)CV²,
where U is the energy stored, C is the capacitance, and V is the voltage across the capacitor.
When a dielectric is inserted between the plates, the electric field between the plates decreases. This is because the dielectric material has a higher permittivity compared to the vacuum. The relationship between the new capacitance (C') with the dielectric and the original capacitance (C) without the dielectric is given by:
C' = KC,
where K is the dielectric constant.
Since the voltage across the capacitor is constant (since the battery has been removed), the change in energy stored can be calculated as:
ΔU = (1/2)C'V² - (1/2)CV²
Substituting the expression for C' and simplifying, we get:
ΔU = (1/2)KC(V² - V²) = 0
Therefore, the change in energy stored is zero, which means there is no change in the energy stored by the parallel-plate capacitor when the dielectric is inserted.
Therefore, the correct answer is:
C) 1 (no change).