Question

What is the strength of the electric field 0.1 mmmm above the center of the top surface of the plate?

Answer 1

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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

Step-by-step explanation:

Solution:

Given:

Diameter of copper = 10cm

Plate charge = 3.5nC = 3.5 *10⁻⁹ C

Finding the radius, we have;

r = D/2

= 10/2

= 5cm

= 5/100 = 0.05m

Calculating the area of the plate, we have

A =πr²

= 3.142 * 0.05²

= 3.142 * 0.0025

= 0.007855m²

Calculating the charge density using the formula;

σ = Q/A

= 3.5 * 10⁻⁹ /0.007855

= 4.4558 * 10⁻⁷ C/m²

The strength of the electric field is calculated using the formula;

E = σ/ε₀

Where ε₀ = 8.84 * 10⁻¹²

E = 4.4558 * 10⁻⁷ / 8.84 * 10⁻¹²

= 5.04 * 10⁴N/C

Answer 2

Answer: 7.38 x 10^3 N/c

Step-by-step 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