Answer :
Final Answer:
(a) The charge on the 6.00 pF capacitor is 3.42 nC, and the charge on the 4.00 pF capacitor is 2.28 nC. (b) The voltage across the 6.00 pF capacitor is 342 V, and the voltage across the 4.00 pF capacitor is 228 V.
Explanation:
To find the charge on each capacitor, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Given that the total potential difference is 570 V and the capacitors are in series, the total capacitance [tex](\(C_{\text{total}}\))[/tex] in a series connection is the inverse sum of the inverse capacitances of each capacitor: [tex]\(\frac{1}{{C_{\text{total}}}} = \frac{1}{C_1} + \frac{1}{C_2}\)[/tex]. Solving for [tex]\(C_{\text{total}}\)[/tex], we find [tex]\(C_{\text{total}} = 2.40\)[/tex] pF. Using Q = CV for each capacitor, we get the charges: [tex]\(Q_{\text{6pF}} = 2.40 \times 570 = 3.42\) nC[/tex] and [tex]\(Q_{\text{4pF}} = 2.40 \times 570 = 2.28\) nC.[/tex]
Next, to determine the voltage across each capacitor, we use [tex]\(V = \frac{Q}{C}\)[/tex]. For the 6.00 pF capacitor, [tex]\(V_{\text{6pF}} = \frac{3.42}{6.00} = 342\)[/tex] V, and for the 4.00 pF capacitor, [tex]\(V_{\text{4pF}} = \frac{2.28}{4.00} = 228\)[/tex] V.
In a series connection, the total charge is the same for each capacitor, while the voltage across each depends on their individual capacitances. The larger capacitor carries a smaller voltage, as observed with the 6.00 pF capacitor having a higher charge but a lower voltage compared to the 4.00 pF capacitor due to their differing capacitances within the series circuit.
