A thin, horizontal, 11-cm-diameter copper plate is charged t

To find the electric field strength ($E$) and direction at a point 0.1 mm above the center of the top surface of the plate, we can use Gauss's law.
The electric field due to a uniformly charged thin plate is given by:
$E=\frac{\sigma }{2{ϵ}_0}$
where $\sigma$ is the surface charge density and ${ϵ}_0$ is the permittivity of free space.
Given that the plate is charged to $-2.9$ nC and has a diameter of 11 cm, we can calculate the surface charge density:
$\sigma =\frac{Q}{A}$ where $Q$ is the charge on the plate and $A$ is the area of the plate. The area of the plate can be calculated using its diameter:
$A=\pi r^2$
Substituting the values, we have:
$\sigma =\frac{-2.9\times {10}^{-9}\text{C}}{\pi {\left(\frac{11\text{cm}}{2}\right)}^2}$
Now, we can substitute the value of $\sigma$ into the formula for electric field $E$ to find its strength:
$E=\frac{\sigma }{2{ϵ}_0}$
To find the direction of the electric field, we use the right-hand rule. If the plate is negatively charged, the electric field points towards the plate.
To calculate the electric field at the plates, we can use the same formula as above:
$E=\frac{\sigma }{2{ϵ}_0}$
However, the direction of the electric field at the plates would be perpendicular to the plates, pointing away from the surface.

Step 1: Find the surface charge density ($\sigma$) of the copper plate.
The surface charge density is defined as the charge per unit area. In this case, the charge is given as -2.9 nC, and the area of the plate can be calculated using the diameter.
The radius ($r$) of the plate is half the diameter, so $r=\frac{11\text{cm}}{2}=5.5\text{cm}$.
The area of the plate ($A$) is given by the formula for the area of a circle: $A=\pi r^2$.
Substituting the values, we have $A=\pi \times (5.5\text{cm}{)}^2$.
Now, we can calculate the surface charge density using the formula $\sigma =\frac{Q}{A}$, where $Q$ is the charge.
Substituting the values, we have $\sigma =\frac{-2.9\text{nC}}{\pi \times (5.5\text{cm}{)}^2}$.
Step 2: Find the electric field strength and direction at a point 0.1 mm above the center of the top surface of the plate.
To find the electric field at a point above the plate, we can use the formula for the electric field due to a uniformly charged infinite plane:
$E=\frac{\sigma }{2{ϵ}_0}$, where $E$ is the electric field strength and ${ϵ}_0$ is the permittivity of free space.
Substituting the value of $\sigma$ into the formula, we have $E=\frac{\frac{-2.9\text{nC}}{\pi \times (5.5\text{cm}{)}^2}}{2{ϵ}_0}$.
To find the direction of the electric field, we consider that it points away from a negatively charged plate.
Step 3: Find the electric field strength between the plates.
The electric field strength between the plates can be calculated using the same formula as in Step 2.
$E=\frac{\sigma }{2{ϵ}_0}$.
Substituting the value of $\sigma$ into the formula, we have $E=\frac{\frac{-2.9\text{nC}}{\pi \times (5.5\text{cm}{)}^2}}{2{ϵ}_0}$.
Since the plates are parallel, the electric field between them is uniform and directed from the positively charged plate to the negatively charged plate.
In summary:
- The surface charge density ($\sigma$) is calculated as $\sigma =\frac{-2.9\text{nC}}{\pi \times (5.5\text{cm}{)}^2}$.
- The electric field strength ($E$) at a point 0.1 mm above the center of the top surface of the plate is given by $E=\frac{\frac{-2.9\text{nC}}{\pi \times (5.5\text{cm}{)}^2}}{2{ϵ}_0}$, and its direction is away from the plate.
- The electric field strength ($E$) between the plates is given by $E=\frac{\frac{-2.9\text{nC}}{\pi \times (5.5\text{cm}{)}^2}}{2{ϵ}_0}$, and its direction is from the positively charged plate to the negatively charged plate.