Answer:
The potential difference across each is [tex]V = V_1 ''=V_2 '' = 818.45\ V[/tex]
The charge on 2.06 μF is [tex]Q_1 '' = 0.00169 \ C[/tex]
The charge on 4.66 μF is [tex]Q_2 '' = 0.00381 \ C[/tex]
Explanation:
From the question we are told that
The capacitance of the first capacitor is [tex]C_1 = 2.60 \mu F = 2.60*10^{-6} \ F[/tex]
The voltage across first capacitor is [tex]V_1 = 1200 V[/tex]
The voltage across second capacitor is [tex]V_2 = 530 V[/tex]
The capacitance of the first capacitor is [tex]C_2 = 4.66 \mu F = 4.66 *10^{-6} \ F[/tex]
The charge on the first capacitor is
[tex]Q_1 = C_1 V_1[/tex]
So [tex]Q_1 = 2.50 *10^{-6} * 1200[/tex]
[tex]Q_1 = 0.003 C[/tex]
The charge on the second capacitor is
[tex]Q_2 = C_2 V_2[/tex]
So [tex]Q_2 = 4.66 *10^{-6} * 530[/tex]
[tex]Q_2 = 0.0025 \ C[/tex]
The capacitance for the connected capacitors is
[tex]C_T = C_1 +C_2[/tex]
So [tex]C_T = 2.06 \mu F + 4.66 \mu F[/tex]
[tex]C_T = 6.72 \mu F[/tex]
The charge on the connected capacitors is
[tex]Q_T = Q_1 + Q_2[/tex]
So [tex]Q_T = 0.003 + 0.0025[/tex]
[tex]Q_T = 0.0055 \ C[/tex]
The voltage across the first capacitor after connection is
[tex]V_1 '' = \frac{Q_T}{C_T}[/tex]
[tex]V_1 '' = \frac{0.0055}{6.72*10^{-6}}[/tex]
[tex]V_1 '' = 818.45\ V[/tex]
The voltage across the second capacitor after connection is
[tex]V_2 '' = \frac{Q_T}{C_T}[/tex]
[tex]V_2 '' = \frac{0.0055}{6.72*10^{-6}}[/tex]
[tex]V_2 '' = 818.45\ V[/tex]
The charge on the first capacitor after connection is
[tex]Q_1 '' = C_1 V_1''[/tex]
So [tex]Q_1 '' = 2.06 *10^{-6} * 818.45[/tex]
[tex]Q_1 '' = 0.00169 \ C[/tex]
The charge on the second capacitor after connection is
[tex]Q_2 '' = C_2 V_2''[/tex]
So [tex]Q_1 '' = 4.66 *10^{-6} * 818.45[/tex]
[tex]Q_2 '' = 0.00381 \ C[/tex]
