A circular capacitor has two metal plates of radius` R=3.0 cm`. The top plate has charge + `Q` , the bottom has charge - `Q` , and they are separated by a short distance. The capacitor is connected to a circuit with current `I=2.5 A` . Find the magnetic field strength `B` at a point between the plates a distance `r=2.0 cm` from the axis through the center of the plates.

Expert Answers

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We can find `B` from the generalized form of Ampere's law. Choose a circular path of radius `r=2.0 cm` about the center line joining the two plates, as shown in the diagram below.

Ampere's law states

`oint_C B^(->)*dl^->=mu_0 (I+I_d)`

Where the displacement current is `I_d=epsilon_0 (d phi_e)/(dt)`

From symmetry the line integral is just `B` multiplied by the circumference of the circle of radius `r` .

`oint_C B^(->)*dl^(->)=B*(2pi r)`

There are no charges moving through the surface `S` , therefore `I=0` .

`B*(2pi r)=mu_0I+mu_0I_d`

`eq. (1) :-gt` `B*(2pi r)=mu_0epsilon_0 (d phi_e)/(dt)`

The electric flux through `S` equals the product of the uniform field strength `E` and the area `A` of the flat surface `S` bounded by the curve `C` , and `E` is equal to the surface charge `sigma ` over `epsilon_0` .

`phi_e=AE=pir^2E=pir^2 sigma/epsilon_0`

`phi_e=pir^2Q/(epsilon_0 pi R^2)=(Qr^2)/(epsilon_0R^2)`

Now substitute these results into `eq. (1)` .

`B*(2pi r)=mu_0epsilon_0 d/(dt)((Qr^2)/(epsilon_0R^2))`

`B*(2pi r)=(mu_0 r^2)/R^2 ((dQ)/(dt))`

`B=(mu_0 r)/(2 R^2 pi) ((dQ)/(dt))`

`B=(m_0 r)/(2 R^2 pi) I`

`B=(2*10^-7 (T*m)/A)(0.02 m)/(0.03 m)^2 (2.5 A)`

`B=1.11*10^-5 T`

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