In: Physics
A parallel plate capacitor with adjustible plate separation d and adjustible area A is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V Volts. (C is the capacitance and U is the stored energy.) Give all correct answers concerning a parallel-plate capacitor charged by a battery (e.g. B, AC, CDF).
A) After being disconnected from the battery, increasing the area A
will increase C.
B) With the capacitor connected to the battery, increasing the area
A will decrease U.
C) With the capacitor connected to the battery, increasing d
increases U.
D) With the capacitor connected to the battery, increasing the area
A will increase Q.
E) After being disconnected from the battery, increasing d
increases V.
F) With the capacitor connected to the battery, increasing d
increases C.
You need to use a few formulas, one for capacitance, one for
energy in a charged capacitor, and one that relates voltage and
stored charge in a capacitor
Eq 1: C= k*epsilon0*A/d
where:
k = relative dielectric constant of dielectric layer
epsilon0 = permittivity of vacuum
A = area of the plates
d = spacing between plates
Eq. 2: U = 1/2*C*V^2
where:
C= capacitance (in Farads)
V = Voltage
Q = Coulombs
U = energy (in joules)
Eq. 3: Q = C*V
A. Rearranging Eq. 3, V = Q/C. Once disconnected, Q is fixed. From
first formula, increasing A increases C, so V must decrease.
B. The battery will keep capacitor voltage constant. From Eq. 1,
decreasing d increases C. From Eq. 2, this means that U must
increase if V is held constant.
C. From Eq. 2, increasing d decreases C. Being connected to battery
does not matter.
D. From Eq 1, decreasing d increases C. Since V is constant (cap
connected to battery), from Eq. 3, Q must increase.
E. Once we disconnect the capacitor from the battery, U stays
constant - (from Conservation of Energy). Although increasing A
causes V to decrease (see question A above), the energy of the
system stays constant.
F. Similar to question C above. From Eq. 1, increasing A increases
C.