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

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles