A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 3.0-µF capacitor?

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