# Homework Help: Net charge

1. Jan 18, 2014

### MathewsMD

Question:
You’re 1.5 m from a charge distribution whose size is much less than 1 m. You measure an electric field strength of 282 N/C. You move to a distance of 2.0 m, and the field strength becomes 119 N/C. What’s the net charge of the distribution? (Hint: Don’t try to calculate the charge. Determine instead how the field decreases with distance, and from that infer the charge.)

I found another similar thread that was closed, and I still had questions.

In terms of finding the net charge, I am slightly confused on how to proceed if it's unnecessary to calculate the charge itself. I have tried, though, and am not getting the correct solution. The answer is 0 from my textbook if anyone is interested, but I can't seem to come up with any mathematical explanation for the answer. From the linked thread, the last person comments on monopoles, dipoles, etc. I don't quite understand how that would allow us to determine the exact net charge given the data from the question. Any help would be greatly appreciated!

2. Jan 18, 2014

### MathewsMD

Also, another question on the same topic:

A thin rod lies on the x-axis between x = 0 and x = L and carries total charge Q distributed uniformly over its length. Show that the electric field strength for x > L is given by $ε = \frac {kQ}{x(x - L)}$

I've done:

$ε = \int_0^Q \frac {k}{(L-x)^2} dq$
$ε = \frac {k}{(L-x)^2} \int_0^Q dq$
$ε = \frac {kQ}{(L-x)^2}$

As you can see this is incorrect. It doesn't seem like an overly difficult problem but I am missing a step somewhere...any guidance on this question would be great too!

3. Jan 18, 2014

### MathewsMD

Also, for the image attached, why would the apparatus not work? Although they all have q/m = a and this will be the only distinguishing factor, since m varies for isotopes, shouldn't it work? It would have to be very precise but I don't quite see why this machine would be dysfunctional.

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4. Jan 18, 2014

### MathewsMD

Sorry for filling this thread with questions. They are all on the same topic of electric fields and I didn't want to start multiple others.

On this problem, I am only an order of magnitude away but can't seem to figure out why...

Find the line charge density on a long wire if a 6.8-μg particle
carrying 2.1 nC describes a circular orbit about the wire with speed 280 m/s.

$ε = \frac {2kλ}{r}$ and $F = qε = \frac {mv^2}{r}$
$\frac {2kλq}{r} = \frac {mv^2}{r}$
$2kλq = mv^2$
$λ = \frac {mv^2}{2kq}$

I simplified this and calculated my answer as 1.41 x 10-3 C/m but the answer is -14 μC/m. I understand it must be negative since it is attractive and I did not include that in my vector for r, but why is the order go magnitude different?

5. Jan 18, 2014

### harts

A dipole has a net charge of zero. A dipole field falls off according to the relationship r-3. If you can show from the data you're given that the field follows this relationship, you can conclude that it is a dipole, and therefore conclude that the net charge is zero.

For the other one, I wouldn't integrate with respect to the charge. I would integrate with respect to x.

6. Jan 18, 2014

### MathewsMD

7. Jan 18, 2014

### harts

As for your confusion on the centripetal force question, notice that 6.8 micrograms is 6.8 E-9 kilograms, not 6.8E-6 kilograms.

8. Jan 18, 2014

### MathewsMD

Thank you for the help with the questions! I'm still doing the integration question but I understood the other 2 now.

I keep getting more questions...

We haven't really discussed structures with more than 2 constituents, so for water, does the dipole moment for each individual O-H bond equal qd still or is it assess differently. Either way, does this question still refer to the fact that $ε ~ \frac {1}{r^3}$...I've been assessing the relationship but don't find anything meaningful...the answer is 0.4e, 0.03 e but i'm not sure why...

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9. Jan 18, 2014

### harts

Notice how it gives the energy that it would take to "reverse the orientation". To go from the lowest potential energy to the highest potential energy, you have ΔU=pE-(-pE).
Check out this wikipedia page - specifically, the section on torque on a dipole. Strictly speaking U is related to the dipole and the electric field by a dot product. The yahoo answer is oversimplistic, like usual.

10. Jan 18, 2014

### MathewsMD

Aha thank you!

For some reason, I decided to consider the initial as 0....:)

Still having some trouble with the other qs...

11. Jan 18, 2014