Question

Consider two oppositely charged, isolated parallel plates separated by distance D, with capacitance C, charge Q, and stored energy U. D is small compared to the dimensions of the plates. For each statement below, select "True" or "False".

1. Because of energy conservation, inserting a dielectric leaves U unchanged.
2. When D is halved, Q stays the same.
3. Inserting a dielectric decreases C.
4. When D is doubled, U increases.
5. When D is doubled, C is doubled.
6. Inserting a dielectric increases Q.
7. Increasing D, leaves the E field unchanged.

Answer 1

Which statements are true for two oppositely charged, isolated parallel plates: C=capacitance, U=stored energy (Q and -Q = charge on the plates). Note: Isolated plates can not lose their charge.

E = ½CV²
Q = CV


When the distance is doubled, C increases.
false, C decreases

Inserting a dielectric decreases U.
inserting a dielectric increases C, which lowers V, since the charge can't change. E = ½CV². Since the V term is squared, the net energy goes down.

check, assume dielectric constant doubles. C then doubles.
by Q = CV, voltage halfs
E = ½CV²
E '= ½(2C)(V/2)² = ½(½)(C)(V)²
E goes down, statement is true

Inserting a dielectric increases C.
true, see above

When the distance is halved, Q stays the same.
Q cannot change
true

When the distance is doubled, U increases.
C = e_0*e_r(A/d)
ε₀ is 8.8542e-12 F/m
εr is dielectric constant (vacuum = 1)
A and d are area of plate in m² and separation in m

double distance, C is cut in half.
cut C in half, V doubles
E = ½CV²
halfing C cuts E in half, doubling V increases E by 4, so net is a doubling in energy. True

Inserting a dielectric increases Q.
charge doesn't change
false

Increasing the distance increases the Electric field
capacitance goes down, charge stays constant, voltage goes up.
Field is V/m, with both voltage and distance going up, so field stays the same.
false