Figure shows a capacitor made of two circular plates each of radius \(12\;\text{cm}\) , and separated by \(5.0\;\text{cm.}\) The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to \(0.15\;A\) .
Calculate the capacitance and the rate of charge of potential difference between the plates.
Calculate the capacitance and the rate of charge of potential difference between the plates.
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Figure shows a capacitor made of two circular plates each of radius \(12\;\text{cm}\) , and separated by \(5.0\;\text{cm.}\) The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to \(0.15\;A\) .
Calculate the capacitance and the rate of charge of potential difference between the plates.
Calculate the capacitance and the rate of charge of potential difference between the plates.
Answer
Distance between the capacitor plates, \(d=5\;\text{cm}=0.05m\)
Radius of circular plate, \(r=12\;\text{cm}=0.12\;m\)
Constant charging current, \(I=0.15A\)
Permittivity of free space, \(\epsilon _o=8.85\times 10^{-12}C^2N^{-1}m^{-2}\)
Capacitance between the plates is given by
\(C=\dfrac{\mathit{\epsilon_o A}} d\)
Where,
\(A=\) Area of each circular plate \(=\pi r^2.\)
\(C=\dfrac{\epsilon _o\pi r^2} d\)
= \(\dfrac{8.85\times 10^{-12}\times \pi \times (0.12)^2}{0.05}\)
\(=8.0032\times 10^{-12}F=80.032\mathit{pF}\)
Charge on each plate, \(q=\mathit{CV}\)
Where,
\(V\) Potential difference across the plates
Differentiating it with respect to time \((t)\) gives:
\(\dfrac{\mathit{dq}}{\mathit{dt}}=C\dfrac{\mathit{dV}}{\mathit{dt}}\)
But, \(\dfrac{\mathit{dq}}{\mathit{dt}}\) = current \((I)\)
\(\dfrac{\mathit{dV}}{\mathit{dt}}\) = \(\dfrac I C\)
⇒ \(\dfrac{0.15}{80.032\times 10^{-12}}\) = \(1.87 × 10^9\;V / s\)
Therefore, the change in potential difference between the plates is \(1.87\times 10^9\ V/\mathit{s.}\)
Radius of circular plate, \(r=12\;\text{cm}=0.12\;m\)
Constant charging current, \(I=0.15A\)
Permittivity of free space, \(\epsilon _o=8.85\times 10^{-12}C^2N^{-1}m^{-2}\)
Capacitance between the plates is given by
\(C=\dfrac{\mathit{\epsilon_o A}} d\)
Where,
\(A=\) Area of each circular plate \(=\pi r^2.\)
\(C=\dfrac{\epsilon _o\pi r^2} d\)
= \(\dfrac{8.85\times 10^{-12}\times \pi \times (0.12)^2}{0.05}\)
\(=8.0032\times 10^{-12}F=80.032\mathit{pF}\)
Charge on each plate, \(q=\mathit{CV}\)
Where,
\(V\) Potential difference across the plates
Differentiating it with respect to time \((t)\) gives:
\(\dfrac{\mathit{dq}}{\mathit{dt}}=C\dfrac{\mathit{dV}}{\mathit{dt}}\)
But, \(\dfrac{\mathit{dq}}{\mathit{dt}}\) = current \((I)\)
\(\dfrac{\mathit{dV}}{\mathit{dt}}\) = \(\dfrac I C\)
⇒ \(\dfrac{0.15}{80.032\times 10^{-12}}\) = \(1.87 × 10^9\;V / s\)
Therefore, the change in potential difference between the plates is \(1.87\times 10^9\ V/\mathit{s.}\)
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