A potential difference V = 100 V is applied across a capacitor arrangement with capacitances C1 = 10.0 mF, C2 = 5.00 mF, and C3 = 4.00 mF. What are:

a. charge q3.
b. potential difference V3.
c. stored energy U3 for capacitor 3.
d. q1
e. V1
f. U1 for capacitor 1
g. q2
h. V2.
i. U2 for capacitor 2?

Given Information:

Potential difference = V = 100 V

Capacitance C₁ = 10 mF

Capacitance C₂ = 5 mF

Capacitance C₃ = 4 mF

Required Information:

a. Charge q₃

b. Potential difference V₃

c. Stored energy U₃

d. Charge q ₁

e. Potential difference V₁

f. Stored energy U₁

g. Charge q  ₂

h. Potential difference V₂  

i. Stored energy U₂

a. Charge q₃ = 0.4 C

b. Potential difference V₃  = 100 V

c. Stored energy  U₃  = 20 J

d. Charge q ₁  = 0.33 C

e. Potential difference  V₁  = 33 V

f. Stored energy U₁  = 5.445 J

g. Charge q  ₂ = 0.33 C

h. Potential difference V₂  = 66 V

i. Stored energy U₂ = 10.89 J

Explanation:

Please refer to the circuit attached in the diagram

a. Charge q₃

As we know charge in a capacitor is given by

q₃ = C₃V₃

q₃ = 4x10⁻³*100

q₃ = 0.4 C

b. Potential difference V₃

The potential difference V₃  is same as V

V₃  = 100 V

c. Stored energy U₃

Energy stored in a capacitor is given by  

U₃  = ½C₃V₃²

U₃  = ½*4x10⁻³*100²

U₃  = 20 J

d. Charge q ₁

Since capacitor C₁ and C₂ are in series their equivalent capacitance is

Ceq = C₁*C₂/C₁ + C₂

Ceq = 10x10⁻³*5x10⁻³/10x10⁻³ + 5x10⁻³

Ceq = 3.33x10⁻³ F

q ₁ = Ceq*V

q ₁ = 3.33x10⁻³*100

q ₁ = 0.33 C

e. Potential difference V₁

V₁  = q ₁/C₁

V₁  = 0.33/10x10⁻³

V₁  = 33 V

f. Stored energy U₁

U₁  = ½C₁V₁²

U₁  = ½*10x10⁻³*(33)²

U₁  = 5.445 J

g. Charge q  ₂

q₂ = Ceq*V

q₂ = 3.33x10⁻³*100

q₂ = 0.33 C

h. Potential difference V₂  

V₂  = q ₂/C₂

V₂  = 0.33/5x10⁻³

V₂  = 66 V

i. Stored energy U₂

U₂ = ½C₂V₂²

U₂ = ½*5x10⁻³*(66)²

U₂ = 10.89 J

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