Given a 3.00 μF capacitor, a 7.75 μF capacitor, and a 5.00 V battery, find the charge on each capacitor if you connect them in the following ways. (a) in series across the battery 3.00 μF capacitor μC 7.75 μF capacitor μC (b) in parallel across the battery 3.00 μF capacitor μC 7.75 μF capacitor μC.

a) Q1= Q2= 11.75×10^-6Coulombs

b) Q1 =15×10^-6coulombs

Q2 = 38.75×10^-6coulombs

Explanation:

a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as

1/Ct = 1/C1 + 1/C2

Given C1 = 3.00 μF C2 = 7.75μF

1/Ct = 1/3+1/7.73

1/Ct = 0.333+ 0.129

1/Ct = 0.462

Ct = 1/0.462

Ct = 2.35μF

V = 5.00Volts

To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance

Q = 2.35×10^-6× 5

Q = 11.75×10^-6Coulombs

Since same charge flows through a series connected capacitors, therefore Q1= Q2=

11.75×10^-6Coulombs

b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2

C = 3.00 μF + 7.75 μF

C = 10.75 μF

For 3.00 μF capacitance, the charge on it will be Q1 = C1V

Q1 = 3×10^-6 × 5

Q1 =15×10^-6coulombs

For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5

Q2 = 38.75×10^-6coulombs

Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2