A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor `R = 1 M Omega` across a 2V battery (Fig. 8.3). 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) = CV [1 – exp `(–t// tau)`], where the time constant `tau` is equal to CR.)

Explanation

The time constant of the CR circuit is `tau = CR = 10^(-3)` s. Then, we have
`q(t) = CV[l-exp(-t//tau)]`
`= 2 xx 10^(-9) [1-exp (-t//10^(-3)]`
The electric field in between the plates at time t is `A = pi(l)^(2) m^(2)` = area of the plates. Consider now a circular loop of radius [ 1/2) m parallel to the plates passing through P. The magnetic field B at all points on the loop is along the loop and of the same value. Then flux
`E xxpixx1/2^(2)=(piE)/(4)=(q )/(4epsilon_(0))`
The displacement current
Now applying amere maxwell law to the loop we get
or `B=0.74 xx10^(-13)` T