Question

A thin, horizontal, 10-cm-diameter copper plate is charged to $3.5\mathrm{nC}$. If the electrons are uniformly distributed on the surface, what are the strength and direction of the electric field

a. $0.1\mathrm{mm}$above the center of the top surface of the plate?

b. at the plate's center of mass?

c. $0.1\mathrm{mm}$below the center of the bottom surface of the plate?

Short answer

(a). The strength of electric field $0.1\mathrm{mm}$above the centre of the top surface of the plate is $25177.05\mathrm{N}/\mathrm{C}$.

The direction of the electric field $0.1\mathrm{mm}$above the centre of the top surface of the plate is vertically upward.

(b). The strength of electric field at the plate's centre of mass is $0\mathrm{N}/\mathrm{C}$.

(c). The strength of electric field $0.1\mathrm{mm}$below the centre of the top surface of the plate is $25177.05\mathrm{N}/\mathrm{C}$.

The direction of the electric field$0.1\mathrm{mm}$ below the centre of the top surface of the plate is vertically downward.

Step-by-step solution

part(a) step 1: given information

The diameter of the copper plate is $10\mathrm{cm}$and the charge is $3.5\mathrm{nC}$.

part(a) step 1: To determine the strength of electric field 0.1 mm above the centre of the top surface of the plate.

Formula to calculate the electric filed strength is,

$E=\frac{\sigma }{2{\mathrm{E}}_0}\left(\mathrm{I}\right)$

Formula to calculate the electric field density is,

$\sigma =\frac{q}{A}$

Formula to calculate the cross-sectional area of the copper plate is,

$A=\frac{\pi d^2}{4}$

Substitute $\frac{q}{A}$for $\sigma$in the equation (I) to find $E$.

role="math" localid="1650125531634" $$\begin{gathered} E=\frac{\frac{4}{A}}{2{ϵ}_0} \\ =\frac{q}{2A{ϵ}_0} \end{gathered}$$

Substitute $\frac{\pi d^2}{4}$for $A$in the above equation to find $E$.

$E=\frac{q}{2\left(\frac{x^2}{4}\right){ϵ}_0}\text{(II)}$

Substitute $3.5\mathrm{nC}$for $q,10\mathrm{cm}$for $d$and $8.85\times {10}^{-12}{\mathrm{A}}^2{\mathrm{s}}^4/{\mathrm{m}}^3\mathrm{kg}$for ${\in }_0$in the above equation to find $E$.

$E=\frac{(3.5\mathrm{nC})\left(\frac{\mathrm{IC}}{{10}^9\mathrm{mC}}\right)}{2\left(\frac{x{\left((10\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}{4}\right)\left(8.85\times {10}^{-12}{\mathrm{A}}^2{\mathrm{s}}^4/{\mathrm{m}}^3\mathrm{kg}\right)}$

$=25177.05N/C.$

part(a) step 2: To determine the direction of the electric field 0.1 mm above the centre of the top surface of the plate.

The direction of electric field is directed away from the positive charge and towards the negative charge. Since the copper plate is positively charged and the point is located above the top surface of the copper plate, so the direction of the electric filed must be away from the copper plate and towards the electric charge that is vertically upward.

part(b) step 1: To determine the strength and direction of electric field at the plate's centre of mass.

From the equation (II), formula to calculate the electric field strength is,

$E=\frac{q}{2\left(\frac{x_d^2}{4}\right){ϵ}_0}$

From the above equation, it can be seen that the electric field strength is directly proportional to the number of charge at surface of the plate. But in case of closed plate, the charge at the centre of the plate is zero.

Substitute $0\mathrm{C}$for $q$in the above equation to find $E$.

$E=\frac{q}{2\left(\frac{x_d^2}{4}\right){ϵ}_0}$

$=\frac{0\mathrm{C}}{2\left(\frac{\pi d^2}{4}\right){ϵ}_0}$

$=0\mathrm{N}/\mathrm{C}$

So, the electric field strength at the centre of the plate is zero.

part(c) step 1: To determine the strength of electric field 0.1 mm below the centre of the top surface of the plate.

From the equation (II), formula to calculate the electric field strength is,

$\mathit{E}=\frac{q}{2\left(\frac{{\pi }^2}{4}\right){ϵ}_0}$

Substitute $3.5\mathrm{nC}$for $q,10\mathrm{cm}$for $d$and $8.85\times {10}^{-12}{\mathrm{A}}^2{s}^4/{\mathrm{m}}^3\mathrm{kg}$for ${\in }_0$in the above equation to find $E$.

$E=\frac{(3.5\mathrm{nC})\left(\frac{1\mathrm{C}}{{10}^{9\mathrm{CC}}}\right)}{2\left(\frac{\times {\left((10\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}{4}\right)\left(8.85\times {10}^{-12}{\mathrm{A}}^2{\mathrm{s}}^4/{\mathrm{m}}^3\mathrm{kg}\right)}$

role="math" localid="1650126557269" $=25177.05N/C$

part(c) step 2: To determine the direction of the electric field 0.1 mm below the centre of the top surface of the plate.

The direction of electric field is directed away from the positive charge and towards the negative charge. Since the copper plate is positively charged and the point is located below the bottom surface of the copper plate, so the direction of the electric filed must be away from the copper plate and towards the electric charge that is vertically downward.