Question 25.46: In Fig. 25-46, how much charge is stored on the parallel-pla......
In Fig. 25-46, how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which \kappa = 3.00; both capacitors have a plate area of 5.00 × 10^{-3} m² and a plate separation of 2.00 mm.
Each capacitor has 12.0 V across it, so Eq. 25-1 yields the charge values once we know C_1 and C_2. From Eq. 25-9,
q = CV, (25-1)
C=\frac{\varepsilon_0 A}{d} \qquad \text { (parallel-plate capacitor). } (25-9)
C_2=\frac{\varepsilon_0 A}{d}=2.21 \times 10^{-11}\, F ,
and from Eq. 25-27,
C=\kappa \varepsilon_0 \mathscr{L} =\kappa C_{\text {air }} (25-27)
C_1=\frac{\kappa \varepsilon_0 A}{d}=6.64 \times 10^{-11}\, F .
This leads to
q_1=C_1 V_1=8.00 \times 10^{-10} \, C , q_2=C_2 V_2=2.66 \times 10^{-10}\, C .
The addition of these gives the desired result: q_{ tot }=1.06 \times 10^{-9}\, C . Alternatively, the circuit could be reduced to find the q_{tot}.