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.

1 Answer

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  • 3 weeks 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. 

    • 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.

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