Consider a capacitor made of two square plates of side 'l' . The distance between the two plates is 'd'.
a) We insert a dielectric of dielectric constant K>1 and width l, a distance x (as in the diagram). What is the total capacitance of this arrangement? Express your answer in terms of the following variables and constants, if relevant, l, x, d, K and epsilon _0 (enter epsilon_0 for epsilon _0).
b) The capacitor is now connected to battery which provides a difference of potential V_0 across the capacitor. What is the energy stored in the capacitor?
Express your answer in terms of the following variables, if relevant, l, x, d, K,V_0 and epsilon _0 (enter epsilon_0 for \epsilon _0 and V_0 for V_0).
c)While the battery is still connected to the capacitor, we now move the dielectric slab a bit further in between the plates, increasing x by an amount Delta. What is the change in the energy stored in the capacitor?
Express your answer in terms of the following variables, if relevant, l, x, d, K, V_0,Delta and epsilon _0 (enter epsilon_0 for epsilon _0 , Delta for Delta and V_0 for V_0).
d) What is the work done by the battery while we push the dielectric slab in from its original position x to x+Delta? (make sure you have the correct sign!) Express your answer in terms of the following variables, if relevant, l, x, d ,K, V_0, Delta and epsilon _0 (enter epsilon_0 for epsilon _0 , Delta for Delta and V_0 for V_0).
e) What is the work done by us while we push the dielectric slab in? (make sure you have the correct sign!) Express your answer in terms of the following variables, if relevant, l, x, d, K, V_0, Delta and epsilon _0 (enter epsilon_0 for epsilon _0 , Delta for Delta and V_0 for V_0).
a) To find the total capacitance of the arrangement, we need to consider the capacitance of the capacitor with and without the dielectric.
Without the dielectric, the capacitance of a parallel-plate capacitor can be calculated using the formula:
C₀ = (epsilon₀ * A) / d
Where:
C₀ = capacitance without dielectric
epsilon₀ = permittivity of free space (a constant)
A = area of one plate (l²)
d = distance between the plates
Therefore, C₀ = (epsilon₀ * l²) / d
When a dielectric is inserted between the plates, the capacitance increases by a factor of the dielectric constant, K. So the total capacitance of the arrangement is:
C = K * C₀
Substituting the value of C₀, we get:
C = K * (epsilon₀ * l²) / d
b) The energy stored in a capacitor is given by the formula:
U = (1/2) * C * V²
Where:
U = energy stored in the capacitor
C = capacitance
V = potential difference across the capacitor
Substituting the values, the energy stored in the capacitor is:
U = (1/2) * (K * (epsilon₀ * l²) / d) * V₀²
= (1/2) * K * epsilon₀ * l² * V₀² / d
c) To find the change in energy stored in the capacitor, we need to calculate the energy stored after moving the dielectric and subtract the initial energy stored.
The capacitance after moving the dielectric by a distance Delta is:
C' = K * (epsilon₀ * l²) / (d + x + Delta)
Therefore, the energy stored after moving the dielectric is:
U' = (1/2) * (K * (epsilon₀ * l²) / (d + x + Delta)) * V₀²
The change in energy is then:
Delta U = U' - U
Substituting the values, we get:
Delta U = (1/2) * (K * epsilon₀ * l² * V₀² / (d + x + Delta)) - (1/2) * (K * epsilon₀ * l² * V₀² / d)
d) The work done by the battery while pushing the dielectric slab from x to x + Delta can be found by calculating the difference in energy stored in the capacitor.
The work done is given by:
W = Delta U = U' - U = (1/2) * (K * epsilon₀ * l² * V₀² / (d + x + Delta)) - (1/2) * (K * epsilon₀ * l² * V₀² / d)
e) The work done by us while pushing the dielectric slab can be found by multiplying the force exerted on the dielectric by the displacement Delta. The force can be calculated using the formula:
F = (1/2) * (A * epsilon₀ * E²)
Where:
F = force exerted on the dielectric
A = area of one plate
epsilon₀ = permittivity of free space (a constant)
E = electric field strength
The electric field strength can be calculated using:
E = V₀ / (d + x + Delta)
Therefore, the work done is:
W = F * Delta = (1/2) * (A * epsilon₀ * E²) * Delta
= (1/2) * (l² * epsilon₀ * (V₀ / (d + x + Delta))²) * Delta
Note: The signs for the work done depend on the direction of the force and displacement.