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A solid ball of radius rb has a uniform charge density ρ.
Part A
What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball?
Express your answer in terms of ρ, rb, r, and ϵ0.
Part B
What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball?
Express your answer in terms of ρ, r, rb, and ϵ0.
Part C
Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?
Check all that apply.
Check all that apply.
| E(0)=0. | |
| E(rb)=0. | |
| limr→∞E(r)=0. | |
| The maximum electric field occurs when r=0. | |
| The maximum electric field occurs when r=rb. | |
| The maximum electric field occurs as r→∞. |
Gauss’s law is the concept that must be used to solve the problem.
Part A: First, use Gauss’s Law to determine the magnitude of E(r), at a distance of r>rb the ball’s center. r is the radius for the solid ball. For part B, use Gauss’s Law to determine
The electric field E(r), at a distance of rrb from center of ball. Part C will reveal the truth about the electric field.
The Gauss’s law states that the net flux of an electrical field within a closed surface is proportional to the charge contained. It can be mathematically calculated by
[katex]\oint {\vec E.d\vec A} = \frac{Q}{{{\varepsilon _0}}}[/katex]
So,
[katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]
Here, E is an electric field, A the area, Q the charge and [katex]{\varepsilon _0}[/katex] are the variables.
This article explains the relationship between charge and density.
[katex]Q = \rho V[/katex]
Here, [katex]\rho [/katex]
(A)
The Gauss’s law was given by
[katex]E\left( r \right) = \frac{Q}{{{\varepsilon _0}A}}[/katex]
But, [katex]Q = \rho V[/katex]
[katex]E\left( r \right) = \frac{{\rho V}}{{{\varepsilon _0}A}}[/katex]
Substitute [katex]\frac{4}{3}\pi r_b^3[/katex] here
[katex]E\left( r \right) = \frac{{\rho \left( {\frac{4}{3}\pi r_b^3} \right)}}{{{\varepsilon _0}A}}[/katex]
Substitute [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]0 now
[katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]1
The distance at which the electric field E(r), magnitude is r>r b From the middle of the ball, [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
(B)
The Gauss’s law was given by
[katex]E\left( r \right) = \frac{Q}{{{\varepsilon _0}A}}[/katex]
But, [katex]Q = \rho V[/katex]
[katex]E\left( r \right) = \frac{{\rho V}}{{{\varepsilon _0}A}}[/katex]
Substitute [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]6
[katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]7
Substitute [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]0 now
[katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]9
The distance from which the electric field E(r), at magnitude E(r), is r <r [katex]{\varepsilon _0}[/katex]0 is the distance from the ball’s center.
(C.1)
This equation is:
[katex]{\varepsilon _0}[/katex]1
Substitute 0 in equation [katex]{\varepsilon _0}[/katex]1
[katex]{\varepsilon _0}[/katex]3
So, [katex]{\varepsilon _0}[/katex]4
(C.2)
This equation is:
[katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
Substitute r b For r in equation [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
[katex]{\varepsilon _0}[/katex]7
So, [katex]{\varepsilon _0}[/katex]8
Thus, [katex]{\varepsilon _0}[/katex]9
(C.3)
This equation is:
[katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
Substitute infinity in equation [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
[katex]Q = \rho V[/katex]2
Thus, [katex]Q = \rho V[/katex]3
(C.4)
This equation is:
[katex]{\varepsilon _0}[/katex]1
Substitute 0 in equation [katex]{\varepsilon _0}[/katex]1
[katex]{\varepsilon _0}[/katex]3
So, [katex]{\varepsilon _0}[/katex]4
The maximum electric field is created when r =0 and _=0 are false.
(C.5)
This equation is:
[katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
Substitute r b For r in equation [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
[katex]{\varepsilon _0}[/katex]7
So, [katex]{\varepsilon _0}[/katex]8
The maximum electric field is therefore when r= or are true.
(C.6)
This equation is:
[katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
Substitute infinity in equation [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2
[katex]Q = \rho V[/katex]2
The maximum electric field is r_, which is false. Part A – Ans
The distance at which the electric field E(r), magnitude is r>r b From the middle of the ball, [katex]E = \frac{Q}{{{\varepsilon _0}A}}[/katex]2 Part A is visible
The distance from which the electric field E(r), at magnitude E(r), is r <r [katex]{\varepsilon _0}[/katex]0 Part A.1 is located at b.
This statement is true. Part C.1
This statement is false. Part C.2
This statement is true. Part C.3
This statement is false. Part C.4
This statement is true. Part C.5
This statement is false. Part C.6