# Uniform charge density and electric potential

• fight_club_alum
In summary, the question asks for the electric potential at a point on the x-axis where a uniformly distributed charge is located from x = a to x = b. The correct answer is c. 49 V. The attempt at a solution involved finding the value of lambda, integrating kq/r from -3 to 2, and using the distance from the point to the charge as the limits of the integral. However, the integral was initially incorrect and should have been from 6 to 11 for 1/x.
fight_club_alum

## Homework Statement

A charge Q is uniformly distributed along the x-axis from x = a to x = b. If Q = 45 nC,
a = –3.0 m, and b = 2.0 m, what is the electric potential (relative to zero at infinity) at the point, x = 8.0 m, on the x axis?

a . 71 V

b. 60 V

c. 49 V <-- correct answer

d. 82 V

e. 150 V

## Homework Equations

lambda = q/x
electric potential = kq/ r

## The Attempt at a Solution

lambda = (45 x 10^=9)/5 = 9 x 10^-9
integrate(kdq/6+x) from -3 to 2
dq = lambda * dx
k*lamda*integral (1/6+x) from -3 to 2 <-- not correct [/B]

fight_club_alum said:

## Homework Statement

A charge Q is uniformly distributed along the x-axis from x = a to x = b. If Q = 45 nC,
a = –3.0 m, and b = 2.0 m, what is the electric potential (relative to zero at infinity) at the point, x = 8.0 m, on the x axis?

a . 71 V

b. 60 V

c. 49 V <-- correct answer

d. 82 V

e. 150 V

## Homework Equations

lambda = q/x
electric potential = kq/ r

## The Attempt at a Solution

lambda = (45 x 10^=9)/5 = 9 x 10^-9
integrate(kdq/6+x) from -3 to 2
dq = lambda * dx
k*lamda*integral (1/6+x) from -3 to 2 <-- not correct [/B]

You need to think about the limits of your integral. Thnik about ##r## = the distance from the point you are interested in (##x = 8m##) and the points where the charge is.

fight_club_alum
fight_club_alum said:
k*lamda*integral (1/6+x) from -3 to 2 <-- not correct
I don't get the integral you are doing. If I read it correctly, you mean
$$\int_{-3}^{2} \frac{1}{6+x} dx$$
which is incorrect. What should ##1/r## be?

Sorry made a stupid mistake should be from integral 6 to 11 (1/x) only

Thanks
fight_club_alum said:
Sorry made a stupid mistake should be from integral 6 to 11 (1/x) only
DrClaude said:
I don't get the integral you are doing. If I read it correctly, you mean
$$\int_{-3}^{2} \frac{1}{6+x} dx$$
which is incorrect. What should ##1/r## be?

## 1. What is uniform charge density?

Uniform charge density refers to a situation where the charge is evenly distributed throughout a given area or volume. This means that the charge per unit volume or surface area is constant.

## 2. How is uniform charge density calculated?

Uniform charge density is calculated by dividing the total charge by the volume or surface area over which it is distributed. This can be represented by the equation ρ = Q/V, where ρ is the charge density, Q is the total charge, and V is the volume or surface area.

## 3. What is the relationship between uniform charge density and electric potential?

The electric potential at a point due to a uniform charge density is directly proportional to the charge density and the distance from the point to the charge. This can be represented by the equation V = kρd, where V is the electric potential, k is a constant, ρ is the charge density, and d is the distance.

## 4. How does a change in uniform charge density affect the electric potential?

An increase in uniform charge density will result in an increase in the electric potential at a given point, while a decrease in charge density will lead to a decrease in electric potential. This is because the electric potential is directly proportional to the charge density.

## 5. Can uniform charge density exist in real-life situations?

Yes, uniform charge density can exist in real-life situations, such as in a parallel plate capacitor where the charge is evenly distributed on the plates. It can also be approximated in situations where the distance between charges is much larger than the size of the charges, such as in a large space with evenly distributed ions.

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