Finding coulmb's force

1. Nov 27, 2007

hastings

[SOLVED] finding coulmb's force

Problem:
A very long straight wire has a uniform linear density charge, λ (every dx, linear unit, has a λ charge). At a distance D, from the wire, there's a rectangular layer with length=a and height=b and a uniform superficial density charge, σ (every dS, superficial unit, has a σ charge). Find the electrostatic force between the 2 objects.

What I did:
(1)$$F=\int dF$$
(2)$$dF=\frac{1}{4\pi \varepsilon_0} \frac{dq_1 \cdot dq_2}{x^2}$$

(3)$$dq_1=\lambda dx$$

(4)$$dq_2=\sigma dS \text{ (* When integrating dS becomes S=a \cdot b)}$$

(5)$$F=\int dF=\int_D^{D+a} \frac{1}{4\pi \varepsilon_0} \frac{(\lambda dx) \cdot \sigma dS }{x^2} = \frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0} \int_D^{D+a} \frac{1}{x^2}dx$$

(6)$$=\frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0}(-\frac{1}{x})_D^{D+a}= \frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0} (\frac{1}{D}-\frac{1}{D+a}) =\frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0}\frac{a}{D(D+a)}$$

(7)$$=\frac{\lambda \sigma (a^2 b)}{4\pi \varepsilon_0 \ D(D+a)}$$

Is my reasoning right? Or, where did I mistake? Please check out the little drawing in the attachment.
Thank you.

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Last edited: Nov 27, 2007
2. Nov 29, 2007

hastings

can anyone help. tell me if it's right or wrong.

3. Nov 29, 2007

Staff: Mentor

No, it's not right.

Don't try to integrate Coulomb's law directly--that's the hard way. (If you do choose that route, you must do it correctly. For one, you must integrate along the full length of the wire and the full area of the rectangle: every element of charge in the wire exerts a force on every element of charge in the rectangle.)

Instead, find the field from the charged wire and figure out the force that that field exerts on an element of charge in your rectangular layer. Then integrate that.

4. Nov 29, 2007

hastings

if I'm not mistaken, the Electrical field at a distance D from the long straight wire should be this,

(2.1) $$E=\frac{\lambda}{2\pi\varepsilon_0D}$$

(2.2) Since $$dF=E\cdot dq_2 \text{ and } dq_2=\sigma$$,

(2.3) $$F=\int dF=\int E dq_2=\int \frac{\lambda}{2\pi\varepsilon_0D}\sigma$$

ok? but when I integrate, I should have something like "dx". Also (a*b) should pop up somewhere. Can't figure out these last two things.

5. Nov 29, 2007

hastings

(3.1) $$F=\int dF=\int E dq_2=\int \frac{\lambda}{2\pi\varepsilon_0D}dq_2=\frac{\lambda}{2\pi\varepsilon_0D}q_2=$$

$$q_2=\sigma(a\cdot b) \longrightarrow \ F=\frac{\lambda}{2\pi\varepsilon_0D} \sigma(a\cdot b)$$

6. Nov 29, 2007

Staff: Mentor

Right. But since D is already used to represent a fixed distance in your problem, better use a variable name such as x. Using x to represent distance from the wire, the field is:

$$E=\frac{\lambda}{2\pi\epsilon_0 x}$$

$$dq = \sigma dA = \sigma b dx$$

The "dx" comes from expressing the charge element in terms of area. The "b" factor comes in when describing the area in terms of length and width. (See what I did directly above.) The "a" factor will come in after you do your integration.

7. Nov 29, 2007

hastings

$$F=\int dF=\int E dq_2=\int_D^{D+a} \frac{\lambda}{2\pi\varepsilon_0x}\sigma b dx$$

is it like this?

When you say $$dq = \sigma dA = \sigma b dx$$

is it like you're dividing the layer into thin vertical, rectangular-shaped strips with height=b, and length=dx?

8. Nov 29, 2007

Staff: Mentor

Yes.

Yes. It makes sense to do it that way, since the field is the same everywhere within such a thin strip.

You could say:
$$dq = \sigma dA = \sigma dy dx$$
But then you'd just have to integrate with respect to y as well. (You'll get the same answer, of course, since the integral of dy will equal b.)

Also, regarding your diagram: Once source of confusion is that you chose the same name (x) for two unrelated things. On the left, you use x to represent position along the line of charge. But you also use x to represent distance from the line charge. Use different variable names!

9. Nov 29, 2007

hastings

Got it. Thank you for all your help.