After leaving a capacitor connected to a battery for a long time, you insert a plastic that has a dielectric K=2. Which statement is true?
a) Q_B = Q_A
b) Q_B = Q_A / 4
c) Q_B = Q_A / 2
d) Q_B = 2Q_A
e) Q_B = 4QA
I am confused as to what will happen when the dielectric is inserted and I keep getting mixed answers. I did C_B = (2 * E naught * A) / d and threw the 2 into the equation.
- billrussell42Lv 72 months ago
Assuming Cb, Qb is after the insertion, Ca, Qa is before
Cb = 2Ca
IF the battery remains connected, (and you don't say) then
Q = CV. Since V is unchanged, when C doubles, then Q doubles.
thus Qb = 2Qa
If the battery is disconnected before the insert
Qb = Qa
since charge is conserved.
in more detail:
Parallel plate cap
C = ε₀εᵣ(A/d) in Farads
ε₀ is vacuum permittivity, 8.854e-12 F/m
εᵣ is dielectric constant or relative permittivity
of the material (vacuum = 1)
A and d are area of plate in m² and separation in m
εᵣ starts at 1 and is changed to 2, so C doubles.