Capacitance — Question 2

Question 2

Signed terminal charges follow a signed voltage convention

Define $v=V_A-V_B$ and $q_A=Cv$ for an isolated ideal capacitor with C=4.00 nF. Initially $V_A=5.00$ V and $V_B=17.0$ V.

1. Find both terminal charges.

2. The voltage is reversed to v=+12.0 V. Find the signed change in charge on A and explain why it exceeds the initial charge magnitude.

Question 2 - Solution

Signed terminal charges follow a signed voltage convention

(a) Method and calculation

$$v=5.00-17.0=-12.0\,\mathrm V,\quad q_A=-48.0\,\mathrm{nC}.$$$$q_B=-q_A=+48.0\,\mathrm{nC}.$$

(b) Interpretation and further calculation

$$q_{A,f}=+48.0\,\mathrm{nC},\quad\Delta q_A=48.0-(-48.0)=96.0\,\mathrm{nC}.$$$$\Delta q_A=C\Delta v=(4.00\,\mathrm{nF})(24.0\,\mathrm V).$$

Physical check

Reversing polarity first neutralizes the original plate charge and then builds the opposite charge.