(a) 4.20 J
(b) 16.74 J
For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;
C = A∈₀ / d --------------------(i)
Where;
∈₀ = constant called permittivity of vacuum.
The energy U stored in such capacitor is given by;
U = [tex]\frac{1}{2}[/tex]CV² ----------------------(ii)
or
U = [tex]\frac{1}{2}[/tex](Q²/C) -------------------(**)
Where;
V = potential difference or voltage across the plates.
Q = charge on the plates.
(a) If the charge is held constant
Combine equations (i) and (**) to give;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d) -----------------------(iii)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iii)
8.40 = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)
8.40 = [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)
Multiply through by 2
2 x 8.40 = Q² x (0.023 / A∈₀)
16.80 = Q² x (0.023 / A∈₀)
Divide through by 0.023
16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023
730.4 = Q² / (A∈₀)
Make Q² subject of the formula
Q² = 730.4(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
Q = constant [this means that Q² still remains 730.4(A∈₀) ]
The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)
U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)
U = 4.20J
Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J
(b) If the voltage is held constant
Combine equations (i) and (ii) to give;
U = [tex]\frac{1}{2}[/tex](A∈₀ / d)V² -----------------------(iv)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iv)
8.40 = [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²
Multiply through by 2 x 0.023
2 x 0.023 x 8.40 = (A∈₀)V²
2 x 0.023 x 8.40 = (A∈₀)V²
0.385 = (A∈₀)V²
Make V² subject of the formula
V² = 0.385/(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
V = constant [this means that V² still remains 0.385/(A∈₀) ]
The energy stored is found by substituting these values of d and V² into equation (iv) as follows;
U = [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]
U = [tex]\frac{1}{2}[/tex](0.385/0.0115)
U = 16.74
Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J
