Consider two oppositely charged, isolated parallel plates separated by distance
ID: 1414334 • Letter: C
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".
True False When D is doubled, U increases.
True False When D is doubled, C is doubled.
True False When D is halved, Q stays the same.
True False Inserting a dielectric increases U.
True False Inserting a dielectric does not affect Q.
True False Because of charge conservation, inserting a dielectric leaves C unchanged.
True False Increasing D, leaves the E field unchanged
Explanation / Answer
When the distance is doubled, U increases.
C = r(A/d)
is 8.8542e-12 F/m
r 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
When D is doubled, C is doubled. -----false, C decreases
When the distance is halved, Q stays the same.......True ,Q cannot change
Inserting a dielectric increases U..........False,
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 false
Inserting a dielectric does not affect Q......... True
Because of charge conservation, inserting a dielectric leaves C unchanged.....False increases C.(see above)
Increasing D, leaves the E field unchanged ....True
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.