A 2.0-μF capacitor and a 4.0-μF capacitor are connected in series across a 1.0-kV potential. The charged capacitors are then disconnected from the source and connected to each other with terminals of like sign together. Find the charge on each capacitor and the voltage across each capacitor.

Explanation:

Given that,

We have two capacitors connected in series

C1=2.0-μF

C2=4.0-μF

Then the equivalent of their series connection

1/Ceq = ½ + ¼

1/Ceq= (2+1)/4

1/Ceq=¾

Taking the reciprocal

Ceq= 4/3 μF

The capacitors are connected to a battery of 1kv

V=1000Volts

We know that,

Q=CV

Where Q is charge

C is capacitance and

V is voltage

Then, Q=4/3 ×1000

Q=4000/3 -μC

Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF

After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,

For parallel connection, they have the same voltage but different charges.

When connected in parallel, there is a charge redistribution,

And the total charge will be 2•4000/3=8000/3 -μF

Then, Q1 +Q2= 8000/3 μF

Now the charge on each capacitor will be, let them have a common voltage V

Q=CV

Then, Q1=C1V

Q1= 2×V=2V

Q2= 4×V=4V

Then, Q1+Q2=8000/3

4V+2V=8000/3

6V=8000/3

V=8000/(3×6)

V=4000/9

V=444.44Volts

Now, Q1=2V

Q1=2×4000/9

Q1=8000/9 μF

Also, Q2=4V

Q2=4×4000/9

Q2=16000/9 μF

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