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 can use the principle of symmetry. Since the plate is thin and has a uniform charge density, the electric field above the center of the top surface will be the same as the electric field directly above the center of the plate.
To calculate the electric field at a point above the center of a uniformly charged disk, we can use the formula:
E = (σ / 2ε₀) * (1 - (z / √(z² + R²)))
Where:
E is the electric field strength
σ is the surface charge density
ε₀ is the permittivity of free space
z is the distance from the center of the disk to the point above it
R is the radius of the disk
Given:
Diameter of the copper plate = 6.00 cm
Radius of the copper plate (R) = 6.00 cm / 2 = 3.00 cm = 0.03 m
Charge on the copper plate (Q) = 6.50 nC = 6.50 × 10⁻⁹ C
Distance from the center of the top surface to the point above it (z) = 0.1 mm = 0.1 × 10⁻³ m
Next, we need to calculate the surface charge density (σ), which is defined as the charge per unit area of the plate. We can use the formula:
σ = Q / A
Where A is the area of the top surface of the plate.
The area of a circle is given by the formula:
A = π * R²
Substituting the values, we have:
σ = (6.50 × 10⁻⁹ C) / (π * (0.03 m)²)
Now, we can substitute the values of σ, R, z, and ε₀ into the formula for electric field:
E = (σ / 2ε₀) * (1 - (z / √(z² + R²)))
Solving the equation will give us the strength of the electric field above the center of the top surface of the copper plate.