Determine the potential of a point of the field created by a metal ball with a surface charge density of 10 to the 11th power C / cm2 and a radius of 1 cm, if the distance from the second point to the surface of the ball is 9 cm
Let’s translate all the values from given to the SI system:
σ = 10 ^ -11 C / cm = 10 ^ -7 C / m
r = 1 cm = 0.01 m.
R1 = 9 cm = 0.09 m.
The potential of the field created by a ball with capacity C and charge q is determined by the following expression:
φ = q / C
The capacity of the ball is determined from the expression:
C = 4π * ε0 * r², where ε0 is the electrical constant 8.85 * 10 ^ -12, r is the radius of the ball.
The field strength is:
E = q / (4π * ε0 * r²)
We have R = r + R1.
Let us express the charge from this expression and substitute it and the capacity into the formula for the field potential:
q = E * 4π * ε0 * r = E * 4π * ε0 * r²
φ = q / C = E * 4π * ε0 * (r + R1) / 4π * ε0 * (r + R1)
Our surface charge density is:
σ = ε0 * E
Taking this into account, the expression for the field potential takes the form:
φ = E * 4π * ε0 * r² / 4π * ε0 * (r + R1) = σ * 4π * r² / 4π * ε0 * (r + R1) = σ * (r + R1) ² / (ε0 * (r + R1))
Substitute the numerical values and determine the field potential:
φ = σ * r² / (ε0 * (r + R1)) = 10 ^ -7 * 0.01² / (8.85 * 10 ^ -12 * 0.1) = 11.3 V.
Answer: potential at 11.3 V.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.