A potential difference of V = 38.0 V is applied across a circuit with capacitances C1 = 4.1 nF, C2 = 4.7 nF, and C3= 2.3 nF, as shown in the figure.

  1. What is the magnitude and sign of q3l, the charge on the left plate of C3 (marked by point A)?

  2. What is the electric potential difference, V3, across C3?

  3. What is the magnitude and sign of the charge q2r, on the right plate of C2 (marked by point B)?



Useful concepts:

Charge for capacitors in series is the same.

Current for resistors in series is the same.

Voltage for resistors in parallel is the same.


$$C_1 + C_2 = C_{12}$$

$$\frac{C_{12} \cdot C_3}{C_{12} + C_3} \cdot ΔV = -6.929×10^{-8} C$$

The way I understand it, is that the Voltage will end up being the same at the end as it returns to the negative side of the Voltage... uhm, source?

But, why do you need to calculate equivalent capacitance? Why can't you just say $C_3 \cdot V_{eq} = Q_3$ ??


$$\frac{Q_3}{C_3} = 30.1 V$$


$$\frac{C_2}{C_1+C_2} \cdot Q_3 = -3.701×10^{-8} C$$

Can someone explain the 3? I don't get the solution... How do you arrive at this equation? $C_1$ and $C_2$ are in parallel, the Charges are divided, but how about the voltage? Is it the same across them?

$$\frac{C_1 \cdot C_2}{C_1+C_2}$$

And if you needed to find V_2, how would you go about it?


1 Answer 1

  1. You do not know $V_{eq}$.

Once you know the PD across C3 you know the voltage across C2 = 38 - 30.1 = 7.9 V

Ans so you can find the charge on C2 = 4.7 x 7.9 = 37.13 nC and then you need to negate this value.

The solution notes that the charge on C3 is the same as the charged on the C1 & C2 parallel combination.

So the voltage across this combination and across one of them C2 is the same so

$V = \dfrac{Q_3}{C_1+C_2} = \dfrac{Q_2}{C_2}$

  • $\begingroup$ What is PD. Also, my question was left unanswered, why can't it be $Q_3 = C_3 \cdot V_{eq}$ $\endgroup$
    – Jack
    Feb 18, 2016 at 3:03
  • $\begingroup$ PD is potential difference or voltage. Q3=C3 V_eq is a valid equation but what is the value of V_eq? $\endgroup$
    – Farcher
    Feb 18, 2016 at 6:05
  • $\begingroup$ the given voltage. Also, how is it valid if the answer to first question (charge $Q_3$) is that you use overall Capacitance, and not just the capacitance of $C_3$ $\endgroup$
    – Jack
    Feb 18, 2016 at 15:28
  • $\begingroup$ The given voltage is not across capacitor C3. The charge on capacitor C3 is the same as the total charge on capacitors C1 and C2. The right hand corner bit of circuit is isolated from the rest of the circuit by the dielectrics with in the capacitors. Before the capacitors were charged I assumed that there was no net charge in that part of the circuit. So for ever more there mast be no net charge. The charges in that region +Q3, -Q1 and - Q2 and they must add to zero, so Q3 = Q1 + Q2. $\endgroup$
    – Farcher
    Feb 18, 2016 at 15:48
  • $\begingroup$ I guess my question is, The formula $Q_3 = C_3 \cdot V_3$ is valid, but how come the formula $Q_3 = C_{eq} \cdot ∆V$ (supplied Voltage by emf?) is also valid?? Because the current has to travel through all the capacitors when it reaches the left plate of $C_3$? Therefore you can use the overall capacitance and the supplied voltage? And treat it as if the circuit only has 1 capacitor? ($C_{eq}$) $\endgroup$
    – Jack
    Feb 19, 2016 at 2:56

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