Question

A parallel plate capacitor with circular plates of radius 1m has a capacitance of 1nF. At t=0, it is connected for charging in series with a resistor R = 1MΩ across a 2V battery. Calculate the magnetic field at a point P, halfway between the centre and the periphery o fthe plates, after t = 10-3 s. (The charge on the capacitor at time t is q(t) = CV[ 1 - exp (-t / τ)], where the time constant τis equal to CR.). Please explain all the calculations.

Answer 1

See the image below...........

Answer 2

The magnetic field at a point P, halfway between the centre and the periphery of the plates is 2.06 × 10⁻¹⁵ T

Charges on the parallel plate capacitor:

The current flowing through the plate at any instant is given by the formula:

$i R+\frac{q}{C}=V$

Where, i is rate of change of electric charges.

$\RIghtarrow i=\frac{d q}{d t}$

Now, the equation becomes,

$R \frac{d q}{d t}+\frac{q}{C}=V$

$R \frac{d q}{d t}=V-\frac{q}{C}$

$R \frac{d q}{d t}=\frac{C V-q}{C}$

Now, on considering CV = q₀, we get,

$R \frac{d q}{d t}=\frac{q_{0}-q}{C}$

$\frac{d q}{q_{0}-q}=\frac{d t}{R C}$

Now, on integrating both sides, we get,

$-\ln \left(q_{0}-q\right)=\frac{t}{R C}+k$

Now, on applying initial condition which is q = 0; t = 0, we get,

$-\ln \left(q_{0}\right)=k$

Now, the equation becomes,

$\ln \left(q_{0}-q\right)-\ln \left(q_{0}\right)=-\frac{t}{R C}$

$\ln \left(\frac{q_{0}-q}{q_{0}}\right)=-\frac{t}{R C}$

$\left(\frac{q_{0}-q}{q_{0}}\right)=e^{-\frac{t}{R C}}$

Now, the equation of charges of plate in series with resistor is:

$\therefore q=q_{0}\left(1-e^{-\frac{t}{R C}}\right)$

Magnetic field at a point P:

Maxwell Ampere equation is given as:

$\int \boldsymbol{B} \cdot \boldsymbol{d} \boldsymbol{l}=\boldsymbol{\mu}_{0} \boldsymbol{i}_{e}+\boldsymbol{\mu}_{0} \varepsilon_{0} \frac{d \Phi_{\boldsymbol{k}}}{d t}$

The electric field between the plates is given by the formula:

$\frac{\sigma}{\epsilon_{0}}=\frac{Q}{\pi R^{2} \epsilon_{0}}$

$\phi_{E}=\frac{Q}{\pi R^{2} \epsilon_{0}} A$

$\because A=\pi r^{2}$

The magnetic field on applying Maxwell theorem, we get,

$\int \boldsymbol{B} \cdot d \boldsymbol{l}=\boldsymbol{B} 2 \pi \mathbf{r}$

$\frac{d \phi_{k}}{d t}=\frac{i\left(\pi r^{2}\right)}{\epsilon_{0}\left(\pi R^{2}\right)}$

$\mathrm{B} 2 \pi \mathrm{r}=\mu_{0} \frac{i\left(\pi \mathrm{r}^{2}\right)}{\left(\pi \mathrm{R}^{2}\right)}$

$\mathbf{B}=\mu_{0} \frac{\text { ir }}{2\left(\pi R^{2}\right)}$

Where, r ≤ R

Now,

$C=\frac{\epsilon_0 A}{d}$

On substituting the values, we get,

$d=\frac{8.85 \times 10^{-12} \times 3.14}{1 \times 10^{-9}}=27.79 \times 10^{-3} \ \mathrm{m}$

$\therefore d \simeq 2.8 \ \mathrm{cm}$

Now,

$q=C V\left[1-e^{-\frac{t}{R C}}\right]$

$\frac{d q}{d t}=\frac{c V}{R C} e^{-\frac{t}{R C}}$

$i=\frac{V}{R} e^{-\frac{i}{RC}}$

On substituting the values, we get,

$i=\frac{2}{1 \times 10^{6}} e^{-\frac{10^{-3}}{10^{6-9}}}$

$\therefore i=0.7357 \times 10^{-6} \ A=7.357 \times 10^{-7} \ A$

Now,

$B=\frac{\mu_{0} i}{2 \pi}\left(\frac{r}{R^{2}}\right)$

On substituting the values, we get,

$B=2 \times 10^{-7} \times 7.357 \times 10^{-7} \times 0.014$

$\therefore B=2.05996 \times 10^{-15} \ T$