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

In summary, the electric field at the point 0.7m from the end of the left rod, which has a charge density of +3 microC/m, is equal to 114997 N/C when combined with the electric field from the right rod, which has a charge density of -4 microC/m. This is derived by integrating the equation dE = (kλdx)/(x+a)^2 from 0 to L and adding the electric fields together. However, the answer may be marked incorrect if there are too few significant figures.
  • #1
Jaccobtw
163
32
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
 
Physics news on Phys.org
  • #2
Here's the problem
 

Attachments

  • Screenshot (92).png
    Screenshot (92).png
    8.9 KB · Views: 173
  • #3
My final answer with symbols
 

Attachments

  • IMG-0755.jpg
    IMG-0755.jpg
    46.2 KB · Views: 169
  • #4
Jaccobtw said:
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.
 
  • Like
Likes Jaccobtw
  • #5
haruspex said:
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.
 
  • Like
Likes SolarisOne and Jaccobtw
  • #6
haruspex said:
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
 
Last edited:
  • #7
Jaccobtw said:
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.
kuruman said:
one cannot go wrong if one specifies too many sig figs
Unless the algorithm checks that.
 
  • #8
kuruman said:
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?
 
  • #9
Jaccobtw said:
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.
 
  • Like
Likes Jaccobtw
  • #10
haruspex said:
I get the same answer as you do.
So do I.
 
  • Like
Likes Jaccobtw
  • #11
haruspex said:
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.
 
  • #12
Charles Teoh said:
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.
 

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

What is an electric field?

An electric field is a physical property that describes the influence of electrically charged objects on each other. It is a vector quantity, meaning it has both magnitude and direction.

How is the electric field calculated?

The electric field at a point can be calculated by dividing the force exerted on a positive test charge at that point by the magnitude of the test charge. It is also dependent on the distance between the point and the charged objects.

What is the difference between a point charge and a charged rod?

A point charge is a theoretical concept of a single, isolated charge with no physical dimension. A charged rod, on the other hand, has a physical shape and can have a distribution of charges along its length.

Can the electric field at a point be negative?

Yes, the electric field at a point can be negative. This indicates that the direction of the electric field is opposite to the direction of the force that would be exerted on a positive test charge at that point.

How does the electric field change as you move away from two charged rods?

The electric field decreases as you move away from two charged rods. This is because the force exerted by the charged rods on a test charge decreases as the distance between them increases, resulting in a weaker electric field.

Similar threads

Back
Top