Question

A square metal plate of edge length 8.0cmand negligible thickness has a total charge of$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{C}$. (a) Estimate the magnitude Eof the electric field just off the center of the plate (at, say, a distance of0.50mmfrom the center) by assuming that the charge is spread uniformly over the two faces of the plate. (b) Estimate Eat a distance of 30m(large relative to the plate size) by assuming that the plate is a charged particle.

Short answer

1. The magnitude of the electric field at $d=0.5mm$is $5.3\times {10}^{7}N/C$.
2. The magnitude of the electric field at $d=30m$is $60N/C$.

Step-by-step solution

The given data

1. The edge length of the square, $a=8cm\left(\frac{1m}{100cm}\right)=0.08m$
2. The total charge on the plate, $q=6\times {10}^{-6}C$
3. Distance of the point from the center,$d=0.5mm\left(\frac{1m}{1000mm}\right)=5.0\times {10}^{-4}m$

Understanding the concept of the electric field

In this case, the charge distributed between the two plates of the sheet can be calculated, Thus, using this value of charge density in the electric field equation of Gauss's flux theorem, we can get the required electric field. Again, for an electric field at a distance, the value can be calculated using the basic formula.

Formulae:

The electric field of a point due to charge particle, $E=\frac{kq}{r^2}$ (1)

The electric field of a conducting sheet,$E=\frac{\sigma }{{\epsilon }_0}$ (2)

The surface charge density of a material, $\sigma =\frac{q}{A}$ (3)

a) Calculation of the electric field at d = 0.5 mm

The charge is distributed uniformly over both sides of the original plate, with half being

on the side near the field point. Thus, the surface density of each plate can be calculated using equation (3) is given as:

$$\begin{gathered} \sigma =\frac{6.00\times {10}^{-6}C}{2(0.08m^2} \\ =4.69\times {10}^{-4}C/m^2 \end{gathered}$$

The magnitude of the field is given using the given data in equation (2) as follows:

$$\begin{gathered} E=\frac{4.69\times {10}^{-4}C/m^2}{8.85\times {10}^{-12}{C}^2/N.m^2} \\ =5.3\times {10}^{7}N/C \end{gathered}$$

The field is normal to the plate and since the charge on the plate is positive, it points away from the plate.

Hence, the magnitude of the electric field is $5,3\times {10}^{7}N/C$ .

b) Calculation of the electric field at d = 30m

At a point far away from the plate, the electric field is nearly that of a point particle with a charge equal to the total charge on the plate. The magnitude of the field is given using equation (1) as follows: (r is the distance from the plate)

$$\begin{gathered} E=\frac{(9\times {10}^{9}N.m^2/C^2)(6.00\times {10}^{-6}C}{{\left(30m\right)}^2} \\ =60N/C \end{gathered}$$

Hence, the value of the electric field is $60N/C$ .