Download Q: A 1140nF capacitor with circular parallel plates 1

Q: A 1140nF capacitor with circular parallel plates 1.12cm in diameter is
accumulating charge at the rate of 20.7mC/s at some instant in time. What will be the
induced magnetic field strength 10.8cm radially outward from the centre of the plates?
What will be the value of the field strength after the capacitor is fully charged?
Solution: C = 1140 nF = 1140 * 10 9 F
d = 1.12cm  0.0112m which gives r = d/2 = 0.0056 m
R = radial distance outside the capacitor = 10.8 cm = 0.108m
dq
 20.7mC / s  20.7 *10 3 C / s (I am assuming
Rate of charge accumulation =
dt
that m stands for milli (10^-3). It might stand for micro (10^-6).

E = , where  = charge density = q/A
0
dE
1 dq


.
dt
A 0 dt
Here dq/dt is given in the question. So rate of change in electric field can be
calculated.
The magnetic field induced outside a circular parallel plate capacitor is given as:
  R 2 dE
B 0 0
2r
dt
Here  0  Permittivity of Free Space = 8.854x10-12 C 2/N.m2
 0  Permeability of Free Space 4π × 10-7 N·A-2
You may now substitute the values in this expression and calculate B.
As soon as the change ion electric field halts, the induced magnetic field vanishes.
Hence when the capacitor is fully charged, B = 0.