Find the electric field at a point away from two charged rods

[Jaccobtw]

Homework Statement

Two rods that are each 1m in length are arranged on an axis so that their ends are 1m apart as shown. The left rod has a charge density λ=+3μC/m and the right rod has a charge density λ=−4μC/m. What is the magnitude of the electric field at the point shown, which is 0.7m from the end of the left rod? Answer in N/C.

Relevant Equations

dE = (kλdx)/(x+a)^2

λ1 = 3 microC/m λ2= -4 microC/m
__________ . __________
l----L1---l-a1-l-a2-l-----L2---l
(Not to scale)

L1 = length of rod 1 (1m)
a1 = length of end of rod 1 to point (0.7m)
L2 = length of rod 2 (1m)
a2 = length of end of rod 2 to point (0.3m)
k = e field constant (9.0e9)

Integrate both sides of dE = (kλdx)/(x+a)^2 from 0 to L

Eventually you get (kλ)(1/a - 1/(a+L))
Because the electric field is positive from the left rod and negative from the right rod, we know the electric field will point to the right at the point so we can add the electric fields together.

((λ1k)(1/a1 - 1/(a1+L1)) + ((λ2k)(1/a2 - 1/(a2+L2))Plug in numbers and add together:

114997 N/C

 

[Jaccobtw]

Here's the problem

 

[Jaccobtw]

My final answer with symbols

 

[haruspex]

Homework Statement:: Two rods that are each 1m in length are arranged on an axis so that their ends are 1m apart as shown. The left rod has a charge density λ=+3μC/m and the right rod has a charge density λ=−4μC/m. What is the magnitude of the electric field at the point shown, which is 0.7m from the end of the left rod? Answer in N/C.
Relevant Equations:: dE = (kλdx)/(x+a)^2

λ1 = 3 microC/m λ2= -4 microC/m
__________ . __________
l----L1---l-a1-l-a2-l-----L2---l
(Not to scale)

L1 = length of rod 1 (1m)
a1 = length of end of rod 1 to point (0.7m)
L2 = length of rod 2 (1m)
a2 = length of end of rod 2 to point (0.3m)
k = e field constant (9.0e9)

Integrate both sides of dE = (kλdx)/(x+a)^2 from 0 to L

Eventually you get (kλ)(1/a - 1/(a+L))
Because the electric field is positive from the left rod and negative from the right rod, we know the electric field will point to the right at the point so we can add the electric fields together.

((λ1k)(1/a1 - 1/(a1+L1)) + ((λ2k)(1/a2 - 1/(a2+L2))Plug in numbers and add together:

114997 N/C

Looks right to me, except that you are quoting an unjustified number of digits. The way the given data are written, you can only justify one significant figure, but I would excuse two.

 

[kuruman]

Looks right to me, except that you are quoting an unjustified number of digits. The way the given data are written, you can only justify one significant figure, but I would excuse two.

I would agree with you 100% if this answer were scored by a human. This one appears to be scored by an algorithm. In such cases there is a percentage tolerance, often set by the instructor, before the algorithm declares an answer incorrect. The (unfortunate) message is that one cannot go wrong if one specifies too many sig figs whilst it is possible for a correctly derived answer to be marked incorrect if there are too few sig figs.

 

[Jaccobtw]

Looks right to me, except that you are quoting an unjustified number of digits. The way the given data are written, you can only justify one significant figure, but I would excuse two.

I tried $$ E = \frac{\lambda k_e}{a} - \frac{\lambda k_e}{a + L}$$ for the integration

 

[haruspex]

I tried $$ E = \frac{\lambda k_e}{a} - \frac{\lambda k_e}{a + L}$$ for the integration

Sorry, I don't understand what that has to do with my comment.

one cannot go wrong if one specifies too many sig figs

Unless the algorithm checks that.

 

[Jaccobtw]

I would agree with you 100% if this answer were scored by a human. This one appears to be scored by an algorithm. In such cases there is a percentage tolerance, often set by the instructor, before the algorithm declares an answer incorrect. The (unfortunate) message is that one cannot go wrong if one specifies too many sig figs whilst it is possible for a correctly derived answer to be marked incorrect if there are too few sig figs.

I'm confused because I got the answer wrong but this is what I used:
$$ E = \frac{\lambda k_e}{a} - \frac{\lambda k_e}{a + L} + \frac{\lambda k_e}{a} - \frac{\lambda k_e}{a + L}$$

I plug in the values for one rod and add it to the values of the other rod. Is this correct?

 

[haruspex]

I'm confused because I got the answer wrong but this is what I used:
$$ E = \frac{\lambda k_e}{a} - \frac{\lambda k_e}{a + L} + \frac{\lambda k_e}{a} - \frac{\lambda k_e}{a + L}$$

I plug in the values for one rod and add it to the values of the other rod. Is this correct?

I get the same answer as you do.
Do you know what the official answer is?

Btw, please don’t use exactly the same name for two different variables, particularly not in the one equation! Use subscripts, as you did in post #1.

 

[kuruman]

I get the same answer as you do.

So do I.

 

[Charles Teoh]

I get the same answer as you do.
Do you know what the official answer is?

Btw, please don’t use exactly the same name for two different variables, particularly not in the one equation! Use subscripts, as you did in post #1.

May I know what is the exact answer for this question? 114997 still incorrect for this question.

 

[haruspex]

May I know what is the exact answer for this question? 114997 still incorrect for this question.

Since @kuruman and I agree with the OP's answer, you can be reasonably sure it is correct.