Electric potential - Is the teacher wrong?

In summary, the problem involves finding the point on the y-axis where the potential is zero, given two point charges placed at the origin and (0m, 0.25m). The solution involves setting up an equation using the distance formula and the equation for electric potential, and solving for y. However, there is a discrepancy between the method used by the student and the method used by the teacher, and further clarification is needed to determine the correct solution.
  • #1
tony873004
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electric potential -- Is the teacher wrong?

Homework Statement


This is from our practice midterm test:

A point charge of 8.0 [tex]\mu C[/tex] is placed at the origin of a coordinate system and a charge of -7.0 [tex]\mu C[/tex] is placed at (0 m, 0.25 m). Find where on the y-axis the potential is zero.


Homework Equations


V=kQ/y


The Attempt at a Solution


[tex]
\begin{array}{l}
d_1 = y \\
d_2 = y - 0.25 \\
\\
\frac{{Q_1 }}{y} = \frac{{Q_2 }}{{y - 0.25}}\,\,\,\, \Rightarrow \,\,\,\,\frac{{Q_1 }}{y} = \frac{{Q_2 }}{{y - 0.25}}\,\,\,\, \Rightarrow \,\,\,\, \\
\\
\,Q_1 \left( {y - 0.25} \right) = yQ_2 \,\,\,\, \Rightarrow \,\,\,\,Q_1 y - 0.25Q_1 = yQ_2 \,\,\,\,\, \Rightarrow \,\,\,\, \\
\\
- 0.25Q_1 = yQ_2 - Q_1 y\,\,\,\, \Rightarrow \,\,\,\, - 0.25Q_1 = y\left( {Q_2 - Q_1 } \right)\,\,\,\, \Rightarrow \\
\\
y = \frac{{ - 0.25Q_1 }}{{\left( {Q_2 - Q_1 } \right)}} = \frac{{ - 0.25\left( {8\mu C} \right)}}{{ - 7\mu C - 8\mu C}} = \frac{{ - 2\mu C}}{{ - 15\mu C}} = \frac{2}{{15}}{\rm{m}} \\
\end{array}
[/tex]

Here's the problem. In class, he said it would boil down to a quadratic, and we should get two answers. On the worksheet, he gives two answers of 2/15, like I got, and 2. But I never got a quadratic. Furthermore, I played around with the online applet for 2 opposite charges, and I don't see the equipotential line for 0 crossing the y-axis more than once. Here's a screen shot:
pm2.GIF


What am I missing here?
 
Last edited:
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  • #2
I'm guessing the teacher is right. In the applet, my charges are equal. It doesn't let me control that. But in the example they are not quite equal, so perpahs that zero line does wrap around. Also, simply plugging (2/15) and (2) into the 8/y = -7/(y-0.25) gives me equal but opposite answers in both cases. But I still don't get a quadratic. My method implies that there are not two answers.
 
  • #3
Yah I think you did something wrong, let me make sure I understand this right...

Q2 is located at the point (m, .25m)?

So you're looking for the charge's distance to the y axis, which for Q1 is simply enough y since it's on the origin, but the distance from Q2 to y is going to be sqrt[m^2+(y-.25m^2)^2] where y is whatever point on the y axis

As for why there's two answers your picture from the applet, while sweet, is implying both charges are sitting on the y axis, which barring typo or me reading it wrong is not the case. Especially if the offset charge is smaller, that "zero line" is going to curve back and touch the y-axis twice
 
  • #4
sorry, its a typo.

charge 2 is at (0 meters, 0.25 meters). Both charges are on the y-axis. The point(s) of interest are also on the y-axis.

Thinking about it, I should probably be doing kq1/y + kq2/(y-0.25) = 0 rather than kq1/y = kq2/(y-0.25). At least that explains the "equal but opposite" dillemma. But I'm still not getting a quadratic.
 
Last edited:
  • #5
Think about it this way: as [itex]y\to\pm\infty[/itex] it begins to look more and more like a charge of [itex]1\mu C[/itex] at the origin. Now envision a test point moving from y = plus infinity toward y = minus infinity. At some point the test point will see the closer but smaller negative charge as exactly balancing out the positive charge at the origin. There are obviously poles at 0.25 m and the origin. Somewhere between these poles the negative charge at 0.25 m and the positive charge at the origin will balance.

In general, the equipotential surfaces for this charge distribution will not be a "quadratic". However, the special case of zero potential is "quadratic" (I put "quadratic" in quotes because the correct term is quadric, not quadratic).
 
  • #6
Yah, note that now with the right form of the equation (V1+V2=0) I'm going to bet 2/15*m, when plugged in doesn't work. Where as with the other way 2m didn't work. I believe that makes sense, explain!
 
  • #7
The potential is zero at 2/15 meters and 2 meters. The potential along the y-axis is quadratic in y.
 
  • #8
I believe you because when I plug them into verify the answers, they work. But how do you get that? I don't get a y^2, I only get a y.
 
  • #9
[tex]\begin{array}{l}
\frac{{kq_1 }}{y} + \frac{{kq_2 }}{{y - 0.25}} = 0 \\
\\
\frac{{kq_1 \left( {y - 0.25} \right) + kq_2 y}}{{y\left( {y - 0.25} \right)}} = 0 \\
\\
kq_1 \left( {y - 0.25} \right) + kq_2 y = 0 \\
\\
kq_1 y - 0.25kq_1 + kq_2 y = 0 \\
\\
kq_1 y + kq_2 y = - 0.25kq_1 \\
\\
y\left( {kq_1 + kq_2 } \right) = - 0.25kq_1 \\
\\
y = \frac{{ - 0.25kq_1 }}{{kq_1 + kq_2 }} \\
\\
y = \frac{{ - 0.25q_1 }}{{q_1 + q_2 }} \\
\end{array}[/tex]

There's no y2, hence only 1 value for y. What am I doing wrong?
 
  • #10
tony873004 said:
What am I doing wrong?
You are dividing by y and y-0.25. You should be dividing by the distance from the point charges. For example, what is the potential at some point away from the y-axis? With point charges [itex]q_1[/itex] and [itex]q_2[/itex] at [itex]\mathbf r_1[/itex] and [itex]\mathbf r_2[/itex], the potential is

[tex]
U(\mathbf r) =
\frac 1{4\pi\epsilon_0}
\left(
\frac {q_1}{||\mathbf r - \mathbf r_1||} +
\frac {q_2}{||\mathbf r - \mathbf r_2||}
\right)
[/tex]

The surface on which the potential is zero is thusly

[tex]
\frac {q_1}{||\mathbf r - \mathbf r_1||} +
\frac {q_2}{||\mathbf r - \mathbf r_2||} = 0
[/tex]

or

[tex]
q_1||\mathbf r - \mathbf r_2|| = -q_2||\mathbf r - \mathbf r_1||
[/tex]

Setting [itex]q_1=8\mu C[/itex], [itex]q_2=-7\mu C[/itex], [itex]\mathbf r_1 = \mathbf 0[/itex], and [itex]\mathbf r_2=0.25\,\text{m}\,\hat y[/itex],

[tex]
8||\mathbf r-0.25\,\text{m}\,\hat y|| = 7 ||\mathbf r||
[/tex]

Squaring to get rid of those vector magnitudes,

[tex]
64 (r^2 - 0.5\,\text{m}\,\mathbf r\cdot\hat y + \frac 1 {16} \,\text{m}^2)
= 49 r^2
[/tex]

For a point on the y-axis ([itex]\mathbf r = y\hat y[/itex]) this becomes

[tex]
64 (y^2 - 0.5\,\text{m}\,y + \frac 1 {16} \,\text{m}^2) = 49 y^2
[/tex]

which simplifies to

[tex]
15 y^2 - 32\,\text{m}\,y + 4 \,\text{m}^2 = 0
[/tex]

This quadratic has solutions

[tex]y=\frac{32\pm28}{30} = \frac 2 {15}, 2[/tex]
 
  • #11
Thanks for that detailed answer! I follow the 1st part, up to where you said "squaring". I'll look at it tommorow. My brain is fried now. Thanks, DH
 
  • #12
The part where you squared r, did you square the vector of y-cap too?

I did it in a weird way, but I got 2/15 and 2. Although I got the answers, I'm not entirely sure it was correct.

I set the 0,0 coor as y and 0, 0.25 as y+0.25
So 8/y + -7/(y+0.25) = 0
8/y = -[-7/(y+0.25)]
8/y = 7/(y+0.25)
y = -2

I also tried 8/y - [-7/(y+0.25)]=0
8/y = [-7/(y+0.25)]
y = -2/15

Can you please explain why the way I used only works when d1=y and d2 = y - 0.25?
And is the way I did it valid?
 
  • #13
tony873004 said:
Thanks for that detailed answer! I follow the 1st part, up to where you said "squaring". I'll look at it tommorow. My brain is fried now. Thanks, DH
Innuendo said:
The part where you squared r, did you square the vector of y-cap too?
Answering both of these questions at once, the Euclidean norm of some vector [itex]\mathbf x[/itex] is, by definition, the number [itex]r[/itex] such that [itex]r*r = \mathbf x \cdot \mathbf x[/itex]. So what if [itex]\mathbf x = \mathbf r - 1/4\,\text{m}\,\hat y[/itex]?

[tex]
\begin{aligned}
||\mathbf r - \frac 1 4 \,\text{m}\,\hat y||^2 &=
(\mathbf r - \frac 1 4 \,\text{m}\,\hat y) \cdot
(\mathbf r - \frac 1 4 \,\text{m}\,\hat y) \\[4pt]
&= r^2 - \frac 1 2 \,\text{m}\,\mathbf r\cdot\hat y + \frac 1 {16} \,\text{m}^2
\end{aligned}
[/tex]

Innuendo said:
I did it in a weird way, but I got 2/15 and 2. Although I got the answers, I'm not entirely sure it was correct.

I set the 0,0 coor as y and 0, 0.25 as y+0.25
So 8/y + -7/(y+0.25) = 0
8/y = -[-7/(y+0.25)]
8/y = 7/(y+0.25)
y = -2

I also tried 8/y - [-7/(y+0.25)]=0
8/y = [-7/(y+0.25)]
y = -2/15

Can you please explain why the way I used only works when d1=y and d2 = y - 0.25?
And is the way I did it valid?
Look at your own post again. You did not get 2/15 and 2. You got -2/15 and -2. So what did you do wrong? The potential due to the charge at the origin is not 8/y. What is the potential at y=-1?

Ignoring off axis points, one can do something similar to what I did above but using absolute values. The potential due to the charge at the origin is 8/|y|, not 8/y. The same goes for the charge at 0.25 meters.

[tex]
\begin{aligned}
& \frac 8 {|y|} + \frac {-7} {|y-0.25|} = 0 \\
& 8 |y-0.25| = 7 |y| \\
& 64 (y-0.5y+0.0625) = 49 y^2 \\
& 15 y^2 - 32y + 4 = 0
\end{aligned}
[/tex]

Which is of course the same quadratic as before (except here I was sloppy with the units).
 
  • #14
I understand the absolute values, but I'm confused about why the second d is y-0.25 instead of y+0.25
I would assume |y| is the first d (the lower coordinate) and |y+0.25| is the second d (the higher coor at 0,0.25) But if you use |y-0.25| wouldn't this coordinate be lower than |y|?
 
  • #15
You use y-0.25 because the second charge is 0.25 meters above the origin. Think about it. Which form describes the distance when the test point y is also located 0.25 meters above the origin: |y-0.25| or |y+0.25| ?
 
  • #16
Oh I see.. I always thought y was used to represent point 1. Didn't know it was used as the test charge.
 

Related to Electric potential - Is the teacher wrong?

What is electric potential?

Electric potential, also known as voltage, is the measure of the potential energy of an electric field at a specific point in space. It is measured in volts (V).

How is electric potential different from electric potential energy?

Electric potential is a measure of the potential energy per unit charge at a specific point in space, while electric potential energy is the total potential energy of a charge in an electric field.

Is electric potential the same as electric potential difference?

No, electric potential and electric potential difference (also known as voltage difference) are not the same. Electric potential difference is the difference in electric potential between two points, while electric potential is the value of the electric potential at a single point.

Why do we use the concept of electric potential?

Electric potential is a useful concept in understanding and analyzing electric fields and their effects on charged particles. It helps us understand the movement and behavior of charges in an electric field.

Is it possible for a teacher to be wrong about electric potential?

Yes, it is possible for a teacher to be wrong about electric potential. Teachers are human and can make mistakes, so it's important to always question and verify information, and to consult multiple sources for accurate information.

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