The final charge on the 3.0-µF capacitor is 1.2 × 10⁻⁵ C
The voltage for the given capacitor is 3.7 V
Explanation:
To find the final charge in the capacitor
We have the formula
Q = CV
Q denotes the charge on a capacitor in Coulombs
C denotes the capacitance in Farads
V denotes the voltage in volts
Let us take Q1 and Q2 has the charge on a capacitor,
Apply the formula here,
Q = CV
Q1 = (3.0 × 10⁻⁶ F)(40 V) = 1.2 × 10⁻⁵ C
The final charge on the 3.0-µF capacitor is 1.2 × 10⁻⁵ C
Q2 = (5.0 × 10⁻⁶ F)(18 V) = 9.0 × 10⁻⁵ C
The final charge on the 5.0-µF capacitor is 9.0 × 10⁻⁵ C
Because the capacitors are connected with the polarity reversed, we subtract the smaller from the larger:
Q1 - Q2 = 3 × 10^-5 C
Finding the total capacitance
Q1 +Q2 = 8.0× 10⁻⁶ C
The total capacitance is 8.0 × 10^-6 F
This will make the voltage:
V= C/Q
V=3×10₋⁵/8×10⁻⁶
V=3.75 V
The voltage for the given capacitor is 3.7 V
