# E = ?

1. Apr 23, 2012

### domyy

1. The problem statement, all variables and given/known data

Find the electric field produced by a 2.5uC point charge at distance of 0.75m.

E = ___ V/m

2. Relevant equations

E = kq/r^2

3. The attempt at a solution

E = (9x10^9 Nm^2/C^2)(25)/(0.75)^2 = 400 x 10^9 V/m

What do you guys think?

2. Apr 23, 2012

### collinsmark

Is the charge q = 2.5 μC, or 25 μC? And what's the 'μ' in 'μC" mean?

3. Apr 23, 2012

### domyy

oh It´s 2.5 and by "u" I meant 'μ'. Thanks.

So, correcting my answer, I will have:

E = (9x10^9 Nm^2/C^2)(2.5 μC)/(0.75m)^2 = 40 x 10^9 V/m

4. Apr 23, 2012

### domyy

Am I supposed to include 10^9 as part of my answer?

Oh, please, help me with this. I need to finish this for tomorrow. I am supposed to use the value I get for E to find PE, I believe, of a 2.5μC charge of opposite sign at the distance 0.75m, I think.

Last edited: Apr 23, 2012
5. Apr 23, 2012

### Pengwuino

No, that's not what he was meaning. What does $\mu$ mean with the SI/base-10 system of units? To give you a hint, what would $2.5mC$ mean? Or $2.5kC$?

6. Apr 23, 2012

### domyy

Well, I know that indicates the charge and that 1μC = 10^6C. Is this what you´re asking?

7. Apr 23, 2012

### Pengwuino

Yes. You didn't include that part in your calculation. That's your problem.

8. Apr 23, 2012

### domyy

Oh nooo. I guess I have several wrong answers then.

Ok, let me see:

If 1μC = 10^-6
Then, 2.5μC = 10^-15?

In that case, = (9x10^9 Nm^2/C^2)(2.5μC x 10^-15)/(0.75)^2 = 22.5/ 0.5625 = 40 x 10^-15 x 10^9 = 40 x 10^-6 V/m

I have 3 problems in which I didn´t include the value 10^-6.

1) Calculate the magnitude of E-field produced by 12μC point charge at a distance of 0.2m.
2) Calculate the force acting on a charge of 10μC placed in an electric field of E = 500V/m.
3) Two positive charges of 4μC each are located at a distance of 10 cm from each other. What´s the magnitude of the force exerted on each other?

In all these cases, I am supposed to include the value for μ, right?

Last edited: Apr 23, 2012
9. Apr 23, 2012

### Pengwuino

What? Where did you get $10^{-15}$?!

10. Apr 23, 2012

### collinsmark

1 μC = 1 x 10-6 C.
(One micro-coulomb equals 1 x 10-6 Coulombs. One micro-coulomb equals one millionth of a Coulomb.)

So 2.5 μC = 2.5 x 10-6 C.

11. Apr 23, 2012

### domyy

Oh I understand.

(9x10^9 Nm^2/C^2)(2.5μC x 10^-6)/(0.75)^2 = 22.5/ 0.5625 = 40 x 10^-6 x 10^9 = 40 x 10^3 V/m

Now if my answer is correct, I am supposed to use this value to find PE of 2.5μC charge of opposite sign at 0.75m.

PE = QED
= (40 x 10^3 V/M)(2.5μC x 10^-6)(0.75) = 75 x 10-3

12. Apr 23, 2012

### Pengwuino

To be clear, $2.5\mu C \times 10^{-6}$ is not correct, it is $2.5 \times 10^{-6}C$.

No. Remember, that formula only works when the electric field along the distance, D, is constant. In this case, moving the other charge in from infinity (the assumption is typically that when the charges are an infinite distance apart, the potential energy is 0) will not give a constant electric field.

The potential energy for a pair of point-charge, brought in from infinity, is given by $PE = {{kq_1 q_2}\over{r}}$

13. Apr 23, 2012

### domyy

It´s said it´s a charge of opposite sign. Does it mean q1 = 2.5 and q2 = -2.5 ?

14. Apr 23, 2012

### Pengwuino

Yes (with the $10^{-6}$).

15. Apr 23, 2012

### collinsmark

Okay, that looks right (the final answer that is, after making note of what Pengwuino points out about the 2.5μC x 10^-6)
It doesn't work out that way in this case. That general approach would work if you are trying to calculate the difference in potential energy of moving a charge within a uniform electric field. But the electric field created by a point charge is not uniform.

If this homework is from a calculus based physics class, you can work out the potential energy formula by evaluating (via dW = F·ds)

$$P.E. = -\int_{\infty}^r k \frac{q_1 q_2}{r^{'2}}dr^'$$

but I'm guessing that this is not a calculus based physics class. If it's not a calculus based physics class, there should be a different formula provided (in your textbook/coursework) which gives you the potential energy of bringing two point charges together. [Edit: See Peng's response. He gives you the formula.]

Last edited: Apr 23, 2012
16. Apr 23, 2012

### domyy

PE = ( 9 x 10^9)(2.5 x 10^-6)(-2.5 x 10^-6)/ (0.75)^2 = 56.25 x 10^-3/ 0.5625 = 100 x 10^-3

I hope this is correct :uhh:

17. Apr 23, 2012

### collinsmark

Look more carefully at the formula that Peng gave in post #12. The fact that the r in the denominator is not squared is not a mistake. (Also, you forgot about a minus sign somewhere.)

[Edit: and where did the "0.5625" come from? :uhh:]

[Another edit: Okay, I just noticed that 0.5625 comes from 0.752. But the overall use still isn't quite right. I suggest just starting from using Pengs equation for PE, given in post 12.]

Last edited: Apr 23, 2012
18. Apr 23, 2012

### domyy

PE = (9 x 10^9)(2.5 x 10^-6)(-2.5 x 10^-6)/ (0.75)^2 = 56.25 x 10^-3/ 0.75 = - 75 x 10^-3

19. Apr 23, 2012

### domyy

All the clarifications that both of you made are really relevant. They will be very helpful while I do the other exercises. I missed small steps while applying the formulas and because of your clarifications, I will go back to previous questions and correct the mistakes.

Thank you SO MUCH!