Answer: 7.38 x 10^3 N/c
Explanation:
The Radius of the Copper Field
R= 10.0cm
The distance of the point from the centre of the plate is
Z = 0.1mm
It shows that Z << R
We can assume that the copper plate is a large uniformly charged disk.
Therefore,
E = σ/2Eo
But,
σ = Q/A
σ = -4.1 x 10^-1 / 3.14 x (0.1)^2
σ = -1.306 x 10^-7 c/m^2
Therefore
E = 1.306 x 10^-7 / 2x 8.85 x 10^-12
E = 7.38 x 10^ 3 N/c
Answer:
Your question is not complete. Below is a similar or complete question to the one you asked.
A thin, horizontal, 10-cm-diameter copper plate is charged to 3.5 nC. If the electrons are uniformly distributed on the surface, what is the strengths of the electric field 0.1 mm above the center of the top surface of the plate?
The strength of the electric field is 5.04 * 10⁴N/C
Explanation:
Solution:
Given:
Diameter of copper = 10cm
Plate charge = 3.5nC = 3.5 *10⁻⁹ C
Finding the radius, we have;
= 10/2
= 5cm
= 5/100 = 0.05m
Calculating the area of the plate, we have
= 3.142 * 0.05²
= 3.142 * 0.0025
= 0.007855m²
Calculating the charge density using the formula;
= 3.5 * 10⁻⁹ /0.007855
= 4.4558 * 10⁻⁷ C/m²
The strength of the electric field is calculated using the formula;
Where ε₀ = 8.84 * 10⁻¹²
E = 4.4558 * 10⁻⁷ / 8.84 * 10⁻¹²
= 5.04 * 10⁴N/C
