Energy Stored in an Electric Field problems and solutions

 Problem #1

What capacitance is required to store an energy of 10 kWh at a potential difference of 1000 V?

Answer;
Known:
Energy stored capacitors is W = 10 kWh = 10 x 10³ x 3600 J = 3.6 X 10⁷ J
The potential difference is V = 1000 V

The energy stored in the capacitor is given by

W = ½ CV²

C = 2W/V² = 2 x 3.6 X 10⁷ J/(1000 V)² = 72 F

Problem #2
How much energy is stored in 1.00 m³ of water due to the “fair weather” electric field of magnitude 150 V/m?

Answer:
Known:
Volume water, Vol = 1.00 m³
electric field of magnitude is E = 150 V/m

The energy stored in the capacitor is given by

W = ½ CV²

With capacitance capacitor is C = ε₀A/d and electric field is E = V/d or V = Ed, then

W = ½(ε₀A/d)(Ed)² = ½(ε₀A/d)(Ed)²

W = ½ ε₀E² (Vol) = ½ (8.85 x 10^-12 C²/Nm²)(150 V/m)²(1.00 m³)

W = 9.96 x 10^-8 J

So in one cubic meter, 9.96 × 10⁻⁸ J of energy are stored

Problem #3
A 2.0 μF capacitor and a 4.0 μF capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitors.

Answer:
Known:
Capacitance capacitors C₁ = 2.0 μF and C₂ = 4.0 μF
Potential difference is V = 300 V

The equivalent capacitance of the C₁ – C₂ combination parallel is:

C_ek = C₁ + C₂ = 6 μF

Then, the energy stored in the capacitor is

W = ½ C_ekV² = ½ (6 x 10^-6 F)(300 V)² = 0.27 J

Problem #4
A parallel-plate air-filled capacitor having area 40 cm² and plate spacing 1.0 mm is charged to a potential difference of 600 V. Find (a) the capacitance, (b) the magnitude of the charge on each plate, (c) the stored energy, (d) the electric field between the plates, and (e) the energy density between the plates.

Answer:
Known:
Parallel plate area, A = 40 cm² = 40 x 10^-4 m²
plate spacing is d = 1.0 mm =  1 x 10^-3 m
potential difference, V = 600 V

(a) capacitance given by

C = ε₀A/d

   = (8.85 x 10^-12 C²/Nm²)(40 x 10^-4 m²)/10^-3 m

C = 3.54 x 10^-11 F = 35.4 pF

(b) magnitude of the charge on each plate is

q  = CV = (3.54 x 10^-11 F)(400 V) = 1,46 x 10^-8 C = 14.6 nC

(c) the stored energy is

W = ½ qV = ½(1,46 x 10^-8 C)(600 V) = 2.83 x 10^-6 J = 2.83 μJ

(d) ) the electric field between the plates is

E = V/d = (400 V)/(1 x 10^-3 m) = 4 x 10⁵ V/m = 0.4 MV/m

(e) the energy density between the plates is

u = W/V

  = W/(Ad) = (2.83 x 10^-6 J)/(40 x 10^-4 m² x 10^-3 m)

u = 0.7075 J/m³

Problem #5
In Fig. 01a, a potential difference V = 100 V is applied across a capacitor arrangement with capacitances C₁  = 10.0 μF, C₂ = 5.00 μF, and C₃ = 15.0 μF. What are (a) charge q₃, (b) potential difference V₃, and (c) stored energy U₃ for capacitor 3, (d) q₁, (e) V₁, and (f) U₁ for capacitor 1, and (g) q₂, (h) V₂, and (i) U₂ for capacitor 2?

Fig.01

Answer:
Capacitors C₁ and C₂ parallel,

C₁₂ = C₁ + C₂ = 10.0 μF + 5.0 μF = 15.0 μF
Capacitors C₁₂ and C₃ series,

C_ek = C₁₂C₃/(C₁₂ + C₃)
 
      = (15.0 μF)(15.0 μF)/(15.0 μF + 15.0 μF)

C_ek = 7.5 μF

(a) charge q₃

Fig.01b capacitors C₁₂ and C₃ stringed series then,

q₃ = q₁₂ = C_ekV = (7.5 μF)(100 V) = 750 μC

(b) potential difference V₃ is

V₃ = q₃/C₃ = 750 μC/15.0 μF = 50 V

(c) stored energy U₃ for capacitor 3 is

U₃ = ½q₃V₃ = ½ (750 μC)(50 V) = 1,88 x 10^-2 J

(d) q₁ is

Potential difference

V₁₂ = q₁₂/C₁₂ = 750 μC/15.0 μF = 50 V

 then,

q₁ = C₁V₁₂ = (10.0 μF)(50 V) = 500 μC

(e) V₁, is

V₁ = V₁₂ = q₁₂/C₁₂ = 750 μC/15.0 μF = 50 V

(f) U₁ for capacitor 1,

U₁ = ½q₁V₁ = ½(500 μC)(50 V) = 1,25 x 10^-2 J

(g) q₂ given by

q₁₂ = q₁ + q₂

q₂ = 750 μC – 500 μF = 250 μF

(h) V₂ given by

V₂ = q₂/C₂ = 250 μF/5.0 μF = 50 V

(i) U₂ for capacitor 2 is

U₁ = ½q₂V₂ = ½(250 μC)(50 V) = 6.25 x 10^-3 J

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