Uniform charge density and electric potential

[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]

 

[PeroK]

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.

 

[DrClaude]

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?

 

[fight_club_alum]

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

 

[fight_club_alum]

Thanks

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

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?