# 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.

Relevance
• 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

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.

• Rendio3 weeks agoReport

Thanks! That was the answer I put first (Qb = 2Qa). And that is literally the question that was on the exam. No mention of disconnecting the battery. The professor mentions disconnecting in another problem.