Question 25.38: In Fig. 25-29, a potential difference V = 100 V is applied a......
In Fig. 25-29, a potential difference V = 100 V is applied across a capacitor arrangement with capacitances C_1 = 10.0 μF, C_2 = 5.00 μF, and C_3= 15.0 μF. What are (a) charge q_3, (b) potential difference V_3, and (c) stored energy U_3 for capacitor 3, (d) q_1, (e) V_1, and (f) U_1 for capacitor 1, and (g) q_2, (h) V_2, and (i) U_2 for capacitor 2?
(a) The potential difference across C_1 (the same as across C_2) is given by
V_1=V_2=\frac{C_3 V}{C_1+C_2+C_3}=\frac{(15.0\, \mu F )(100 \,V )}{10.0 \,\mu F +5.00 \,\mu F +15.0 \,\mu F }=50.0 \,V .
Also, V_3 = V – V_1 = V – V_2 = 100 \,V – 50.0 \,V = 50.0 V. Thus,
\begin{aligned} & q_1=C_1 V_1=(10.0\, \mu F )(50.0\, V )=5.00 \times 10^{-4} C \\ & q_2=C_2 V_2=(5.00 \,\mu F )(50.0 \,V )=2.50 \times 10^{-4} \,C \\ & q_3=q_1+q_2=5.00 \times 10^{-4} \,C +2.50 \times 10^{-4} \,C =7.50 \times 10^{-4} \,C . \end{aligned}
(b) The potential difference V_3 was found in the course of solving for the charges in part (a). Its value is V_3 = 50.0 V.
(c) The energy stored in C_3 is U_3=C_3 V_3^2 / 2=(15.0 \,\mu F )(50.0 \,V )^2 / 2=1.88 \times 10^{-2} \,J .
(d) From part (a), we have q_1=5.00 \times 10^{-4} \,C , and
(e) V_1 = 50.0 V, as shown in (a).
(f) The energy stored in C_1 is U_1=\frac{1}{2} C_1 V_1^2=\frac{1}{2}(10.0 \,\mu F )(50.0\, V )^2=1.25 \times 10^{-2} \,J .
(g) Again, from part (a), q_2=2.50 \times 10^{-4}\, C .
(h) V_2 = 50.0 V, as shown in (a).
(i) The energy stored in C_2 is U_2=\frac{1}{2} C_2 V_2^2=\frac{1}{2}(5.00 \,\mu F )(50.0 \,V )^2=6.25 \times 10^{-3}\, J .