A parallel-plate air capacitor is made from two plates 0.250 m square, spaced 0.70 cm apart. It is connected to a 140 V battery.
a) What is the capacitance?
C = 7.90×10−11 F
b) What is the charge on each plate?
Enter your answer as an absolute value of charge.
Q = 1.11×10−8 C
c) What is the electric field between the plates?
E = 2.00×104 V/m
d)What is the energy stored in the capacitor?
U = 7.74×10−7 J
e) If the battery is disconnected and then the plates are pulled apart to a separation of 1.40 cm , what are the answers to parts A, B, C, and D?
Enter your answer as four numbers corresponding to C, Q, E, U. Please enter the answer in the given order and in the same units as in parts A, B, C, and D.
*I got the answers right for parts a-d, pleas help with part e!!!!
To find the answers to part e, we need to consider the change in the parallel-plate capacitor due to the increase in separation between the plates.
The formula for the capacitance of a parallel-plate capacitor is given by:
C = ε0 * (A / d)
Where:
C = Capacitance
ε0 = Permittivity of free space (8.85 × 10^-12 F/m)
A = Area of each plate
d = Distance between the plates
a) Capacitance (C):
We already calculated this in part a. The answer is C = 7.90×10^-11 F.
b) Charge on each plate (Q):
The charge on each plate can be calculated using the formula:
Q = C * V
Where:
Q = Charge on each plate
C = Capacitance
V = Voltage
Substituting the given values into the formula, we have:
Q = (7.90×10^-11 F) * (140 V) = 1.11×10^-8 C
c) Electric Field (E):
The electric field between the plates of a parallel-plate capacitor can be calculated using the formula:
E = V / d
Where:
E = Electric Field
V = Voltage
d = Distance between the plates
Substituting the given values into the formula, we have:
E = (140 V) / (0.70×10^-2 m) = 2.00×10^4 V/m
d) Energy stored in the capacitor (U):
The energy stored in a capacitor can be calculated using the formula:
U = (1/2) * C * V^2
Where:
U = Energy stored in the capacitor
C = Capacitance
V = Voltage
Substituting the given values into the formula, we have:
U = (1/2) * (7.90×10^-11 F) * (140 V)^2 = 7.74×10^-7 J
Now, for part e, where the separation between the plates is increased to 1.40 cm:
a) Capacitance (C):
Using the same formula as in part a, but with the new separation value (1.40 cm = 0.014 m), we have:
C = ε0 * (A / d) = (8.85×10^-12 F/m) * (0.250 m^2 / 0.014 m) = 1.59×10^-9 F
b) Charge on each plate (Q):
Using the formula from part b, we can determine the charge on each plate with the new capacitance:
Q = C * V = (1.59×10^-9 F) * (140 V) = 2.22×10^-7 C
c) Electric Field (E):
Using the formula from part c, we can determine the electric field with the new separation value:
E = V / d = (140 V) / (0.014 m) = 1.00×10^4 V/m
d) Energy stored in the capacitor (U):
Using the formula from part d, we can determine the energy stored in the capacitor with the new capacitance:
U = (1/2) * C * V^2 = (1/2) * (1.59×10^-9 F) * (140 V)^2 = 1.55×10^-6 J
Therefore, the answers for part e are:
C = 1.59×10^-9 F
Q = 2.22×10^-7 C
E = 1.00×10^4 V/m
U = 1.55×10^-6 J