A 300 V battery is connected across capacitors of 3μF and 6 μF in parallel. calculate the energy stored in each capacitor.
Solution
Emf (V) = 300 V
C₁ = 3μF = [latex]3\times 10^{-6}F[/latex]
C₂ = 6 μF = [latex]6\times 10^{-6}F[/latex]
Energy stored in capacitor 1 (E₁) = [latex]\frac{1}{2}C_1V^2 = \frac{1}{2}\times(3\times 10^{-6})\times(300)^2 = 0.135\ J[/latex]
Energy stored in capacitor 2 (E₂) = [latex]\frac{1}{2}C_2V^2 = \frac{1}{2}\times(6\times 10^{-6})\times(300)^2 = 0.27\ J[/latex]
A parallel plate air capacitor of capacitance [latex]245\times 10^{-12}F[/latex] has a charge of magnitude 0.148 μC on each plate. Find the potential difference and electric field intensity between the plates if the distance between plates is 5 mm.
Solution
Capacitance (C) = [latex]245\times 10^{-12}F[/latex]
Charge (q) = 0.148 μC = [latex]0.148\times 10^{-6}C[/latex]
Potential difference (V) = ?
Electric field intensity (E) = ?
Distance between plates (d) = 5 mm = [latex]5\times 10^{-3}m[/latex]
We know that,
q = CV
Or, [latex]V = \frac{q}{C} = \frac{(0.148\times 10^{-6})}{(245\times 10^{-12})} = 604.082\ V[/latex]
Now,
[latex]E = \frac{V}{d} = \frac{604.082}{5\times 10^{-3}} = 1.2\times 10^5[/latex] V/m
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