Electric Field calculation question

[mikej_moore]

Homework Statement

You're 1.5m from a charge distribution whose size is much less than 1m. You measure an electric field strength of 282 N/C. You move to a distance of 2.0m, and the field strength becomes 119 N/C. What is 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.

Homework Equations

E = k*Q/r²

The Attempt at a Solution

E₂-E₁=k*Q(1/r₂²-1/r₁²)
From this I got Q = 9.3E-8...however when I plug this into the relevant equation I don't get the right answers for the first or second charge distance pair. Also shouldn't the field strengths be different? I thought that the change of electric field was inversely proportional to the change in distance...so if you increase the distance by a factor of 1.33 shouldn't the field drop by a factor of 1.77? Shouldn't this give a value of 159 for the second electric field value?

Any help or guidance is appreciated! :D

 

[kuruman]

You completely ignored the hint "Determine instead how the field decreases with distance, and from that infer the charge." Take the ratio E₁/E₂. Is it equal to the ratio r₂²/r₁²?

 

[mikej_moore]

E₁=282 N/C r₁=1.5 m
E₂=119 N/C r₂=2m

E₁/E₂ = 282/119 = 2.4
r₂²/r₁² = 4/2.25 = 1.78
Shouldn't these two values be equal? That's what I don't understand about the question...or maybe I'm calculating something wrong?

 

[kuruman]

They would be equal if you have just one charge Q. Obviously, that's not what you have. This means that the charge distribution does not give rise to a monopole field. If that's not the case, what else could it be?

 

[gabbagabbahey]

You might find my reply in this thread useful...

P.S. hoycey wouldn't happen to be another account of yours would it?

 

[mikej_moore]

Sorry, I'm not quite sure what a multipole expansion is, but I'm assuming that since we're far away the term that's going to dominate is the distance, right?

Also I found that the correct ratio here is E₁/E₂ = d₂³/d₁³...but I'm not sure how to incorporate that into the question...

(And no, I only have one account on this website)

 

[gabbagabbahey]

The multipole expansion, is basically a Taylor expansion of the field or potential in powers of \frac{1}{r} (where r is the distance from the origin)...

The lowest order term in the expansion corresponds to the monopole moment, and it is proportional to \frac{1}{r^2} (just like the field of a point charge!). The second lowest order term is the dipole term, and it is proportional to \frac{1}{r^3} (just like the field of an ideal dipole!). The next term in the expansion is the quadrapole term (proportional to \frac{1}{r^4})...and so on.

At "large" distances (which means small 1/r), the lowest order non-zero terms will dominate (1/r^4 will be much smaller than 1/r^3 etc.)

You really haven't learned about dipole moments yet?

 

[mikej_moore]

Hmm, so because this electric field decreases with the same proportionality as a dipole...can I assume that like a dipole the net charge over the distribution is zero?

 

[kuruman]

You can. If there were a monopole term too, it will would dominate the dipole term and the spatial dependence of the electric field would not be what it is.