A large parallel-plate capacitor with uniform surface charge $\mathit{\sigma }$on the upper plate and $\mathbf{-}\mathit{\sigma }$on the lower is moving with a constant speed localid="1657691490484" $\mathit{\upsilon }$,as shown in Fig. 5.43.

(a) Find the magnetic field between the plates and also above and below them.

(b) Find the magnetic force per unit area on the upper plate, including its direction.

(c) At what speed $\mathit{\upsilon }$would the magnetic force balance the electrical force?

Write the expression for magnetic field of an infinite uniform surface current.

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{K}}{\mathbf{2}}$ …… (1)

Here,${\mu }_0$is the permeability of the air or free space.

Write the expression for surface current in terms of surface charge density and velocity.

$\mathit{K}\mathbf{=}\mathit{\sigma }\mathit{\upsilon }$ …… (2)

Here,$\sigma$is the surface charge density of the conductor and$\upsilon$is the velocity of each plate.

The following diagram shows a large parallel plate capacitor with surface charge $+\sigma$density for upper plate and $-\sigma$for the lower plate of the capacitor.

a)

From the above figure,

Write the expression for the top plate produces a magnetic field which is out page and its below point shows into page.

$B=\frac{{\mu }_{0}K}{2}$

Write the expression for the bottom plate produces a magnetic field which is into page and its below point shows into page.

$B=\frac{{\mu }_{0}K}{2}$

Therefore, the fields due to the plates cancel for point above the top plate and for points below the bottom plate. The fields add up between the plates.

Write the expression for magnetic field between the plates.

$\begin{array}{c}B=\frac{{\mu }_{0}K}{2}+\frac{{\mu }_{0}K}{2}\\ ={\mu }_{0}K\end{array}$

Substitute$\sigma \upsilon$ for$K$

$B={\mu }_0\sigma \upsilon$ …… (3)

Thus, the magnetic field is $B={\mu }_0\sigma \upsilon$.

b)

Write the expression for the according to Lorentz force acting on the upper plate due to lower plate.

$F=\int (K\times B)da$ …… (4)

Here,$K$ is the surface current and$B$is magnetic field.

Therefore, write the expression the force per unit area.

$f=K\times B$ …… (5)

Substitute$\frac{{\mu }_{0}K}{2}\widehat{y}$ for$B$and$\sigma \upsilon \widehat{x}$ for$K$in equation (5)

$\begin{array}{c}{f}_m=\left(\sigma \upsilon \widehat{x}\right)\times \left(\frac{{\mu }_{0}K}{2}\widehat{y}\right)\\ =\sigma \upsilon \frac{{\mu }_{0}K}{2}(\widehat{x}\times \widehat{y})\\ =\sigma \upsilon \frac{{\mu }_{0}K}{2}\widehat{z}\end{array}$

Substitute$\sigma \upsilon$for$K$in above equation.

$f_m=\frac{{\mu }_0{\left(\sigma \upsilon \right)}^2}{2}$

Thus, the magnetic force per unit area is$f_m=\frac{{\mu }_0{\left(\sigma \upsilon \right)}^2}{2}$and the direction of force is upward.

c)

Write the expression for electric field due to lower plate.

$E=\frac{\sigma }{2{\in }_0}$

Here , $E$is the electric field and ${\in }_0$is the permittivity.

Write the expression for electric force per unit area $f_e$on upper plate.

$f_e=\frac{{\sigma }^2}{2{\in }_0}$

Now, balance electric and magnetic forces,

$\begin{array}{c}{f}_e=f_m\\ \frac{{\sigma }^2}{2{\in }_0}=\frac{{\mu }_0{\sigma }^2{\upsilon }^2}{2}\\ {\upsilon }^2=\frac{1}{{\mu }_0{\in }_0}\\ \upsilon =\frac{1}{\sqrt{{\mu }_0{\in }_0}}\end{array}$

Substitute $4\pi \times {10}^{-7}H/m$for permeability of air free space $\left({\mu }_0\right)$and $8.85\times {10}^{-12}{C}^2/N\cdot m^2$for permittivity of the air or free space.

$\begin{array}{c}\upsilon =\frac{1}{\sqrt{4\pi \times {10}^{-7}\times 8.85\times {10}^{-12}}}m/s\\ =3\times {10}^{8}m/s\end{array}$

Thus, the speed of the upper plate is $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$.