A thin copper plate of diameter 6.00 cm is charged to 6.50 nC. What is the strength of the electric field 0.1 mm above the center of the top surface plate?
To find the strength of the electric field above the center of the top surface of the copper plate, we need to use the formula for the electric field due to a charged disk. The formula is:
E = sigma / (2 * epsilon * z)
Where:
- E is the electric field strength,
- sigma is the surface charge density,
- epsilon is the permittivity of the medium,
- z is the distance from the center of the disk.
In this case, we are given the charge of the copper plate, but we need to calculate the surface charge density. To do that, we will use the formula:
sigma = Q / A
Where:
- Q is the charge on the copper plate,
- A is the area of the copper plate.
The area of the copper plate can be calculated using the formula for the area of a circle:
A = pi * r^2
Where:
- r is the radius of the copper plate.
In this case, the diameter of the copper plate is given as 6.00 cm, so the radius is half of that, which is 3.00 cm or 0.03 m.
Let's calculate the area of the copper plate:
A = pi * (0.03)^2
A = 0.002827 m^2
Now we can calculate the surface charge density:
sigma = 6.50 nC / 0.002827 m^2
sigma = 2.3019 C/m^2
The permittivity of the medium can be taken as the permittivity of free space, which is approximately:
epsilon = 8.854 x 10^-12 C^2/(N*m^2)
The distance from the center of the disk to the point above it is given as 0.1 mm, which is 0.0001 m.
Now we can substitute all the values into the formula to find the electric field strength:
E = (2.3019 C/m^2) / (2 * (8.854 x 10^-12 C^2/(N*m^2)) * 0.0001 m)
E ≈ 1306370 N/C
Therefore, the strength of the electric field 0.1 mm above the center of the top surface plate is approximately 1,306,370 N/C.