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## QuestionPhysicsClass 12

A parallel plate capacitor with circular plates of radius $$1\; m$$ has a capacitance of $$1\;\mathit{nF}$$ . At $$t=0$$ , it is connected for charging in series with a resistor $$R=1\;\mathit{M{\Omega}}$$  across a $$2\;V$$  battery . Calculate the magnetic field at a point P, halfway between the centre and the periphery of the plates, after $$t = 10^{–3} s$$. $$($$The charge on the capacitor at time t is $$q(t)=\mathit{CV}[1\text{-}e^{\text{-}\frac t{\tau }}]$$ , where the time constant τ is equal to $$CR).$$

Here the time constant is given by $$\mathit{CR}=10^{-3}\mathit{sec}$$
And the charge on the capacitor at any time t is given by $$q(t)=\mathit{CV}[1-\exp \left(\frac{-t}{\tau }\right)]$$
= $$2\times 10^{-9}[1-\exp \left(\frac{-t}{\tau }\right)]$$
We also know that
Electric field in between the plates
$$E=\dfrac{q}{A \epsilon_0}$$
Here area= $$\pi r^2,$$ $$r=1$$
E= $$\frac q{\mathit{\pi \epsilon_0}}$$
Consider a loop of radius ½ parallel to the plated of the capacitor. So by gauss law
Electric flux $$=E\times A$$
= $$\frac{\mathit{E\pi }} 4=\frac q{4\mathit{\upsilon o}}$$
Maxwell developed displacement current which is
Id= $$\frac{\mathit{\epsilon_0 d\varphi }}{\mathit{dt}}=0.5\times 10^{-5}\exp \left(-1\right)$$
A t$$=10^{-3}\;\text{sec}$$
Applying the ampere maxwell law to calculate the magnetic field
$$B\times 2\pi \times \frac 1 2=\mathit{\mu o}\left(\mathit{ic}+\mathit{id}\right)$$
so we get $$B=0.74\times 10^{-13}T$$          