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Question: Find the energy of a sphere capacitor of radius 2 cm carrying a charge 5µC...

Find the energy of a sphere capacitor of radius 2 cm carrying a charge 5µC

A

5.6 x 10⁻¹ J

B

5.6 J

C

1.39 x 10⁻¹⁹ J

D

0.39 J

Answer

5.6 J

Explanation

Solution

The energy of a charged capacitor is given by the formula U=12CV2=12Q2C=12QVU = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}QV.

A single isolated conducting sphere of radius RR carrying a charge QQ can be considered as a capacitor where the other plate is at infinity.

The capacitance of an isolated conducting sphere of radius RR is given by C=4πε0RC = 4\pi\varepsilon_0 R.

Given:

  • Charge on the sphere, Q=5 μC=5×10−6 CQ = 5 \, \mu\text{C} = 5 \times 10^{-6} \, \text{C}
  • Radius of the sphere, R=2 cm=2×10−2 mR = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}
  • Permittivity of free space, ε0≈8.854×10−12 F/m\varepsilon_0 \approx 8.854 \times 10^{-12} \, \text{F/m}

It is common to use the value of 14πε0≈9×109 Nm2/C2\frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9 \, \text{Nm}^2/\text{C}^2.

Using the formula U=12Q2CU = \frac{1}{2}\frac{Q^2}{C}:

First, calculate the capacitance C=4πε0RC = 4\pi\varepsilon_0 R. We can write C=R1/(4πε0)C = \frac{R}{1/(4\pi\varepsilon_0)}. Using 14πε0≈9×109 Nm2/C2\frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9 \, \text{Nm}^2/\text{C}^2:

C=2×10−2 m9×109 Nm2/C2=29×10−11 FC = \frac{2 \times 10^{-2} \, \text{m}}{9 \times 10^9 \, \text{Nm}^2/\text{C}^2} = \frac{2}{9} \times 10^{-11} \, \text{F}

Now, calculate the energy UU:

U=12Q2C=12(5×10−6 C)229×10−11 FU = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} \frac{(5 \times 10^{-6} \, \text{C})^2}{\frac{2}{9} \times 10^{-11} \, \text{F}} U=1225×10−122/9×10−11 J=12×25×10−122/9×10−11=25×94×10−1210−11U = \frac{1}{2} \frac{25 \times 10^{-12}}{2/9 \times 10^{-11}} \, \text{J} = \frac{1}{2} \times \frac{25 \times 10^{-12}}{2/9 \times 10^{-11}} = \frac{25 \times 9}{4} \times \frac{10^{-12}}{10^{-11}} U=2254×10−1=56.25×0.1=5.625 JU = \frac{225}{4} \times 10^{-1} = 56.25 \times 0.1 = 5.625 \, \text{J}

Alternatively, calculate the potential on the surface of the sphere V=14πε0QRV = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}.

V=(9×109 Nm2/C2)×5×10−6 C2×10−2 m=9×109×5×10−62×10−2=9×109×2.5×10−4=22.5×105 VV = (9 \times 10^9 \, \text{Nm}^2/\text{C}^2) \times \frac{5 \times 10^{-6} \, \text{C}}{2 \times 10^{-2} \, \text{m}} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{2 \times 10^{-2}} = 9 \times 10^9 \times 2.5 \times 10^{-4} = 22.5 \times 10^5 \, \text{V}

Using the formula U=12QVU = \frac{1}{2}QV:

U=12×(5×10−6 C)×(22.5×105 V)=12×5×22.5×10−6×105=12×112.5×10−1=56.25×10−1=5.625 JU = \frac{1}{2} \times (5 \times 10^{-6} \, \text{C}) \times (22.5 \times 10^5 \, \text{V}) = \frac{1}{2} \times 5 \times 22.5 \times 10^{-6} \times 10^5 = \frac{1}{2} \times 112.5 \times 10^{-1} = 56.25 \times 10^{-1} = 5.625 \, \text{J}

The calculated energy is approximately 5.6 J.