# Electric Charges

1. Dec 2, 2003

### marshall4

Charge A is +2.5*10^-5C Charge B is -3.7*10^-7C.
A is 25cm away from B. Point C is 10cm away from point B. What is the electric field at C? All points are in a line A--B-C.

2. Dec 2, 2003

### PrudensOptimus

$$A: 25\mu C, B: -.37\mu C, C: ?$$

$$\begin{equation*} \begin{split} E &= \frac{F}{q} &= \frac{kqQ/r^2}{q} &= k\frac{Q}{r^2} &= \frac{Q}{4\pi \sigma_{0}r^2} \end{split} \end{equation*}$$

Last edited: Dec 2, 2003
3. Dec 2, 2003

### marshall4

I have no idea what you just wrote. What numbrs do i put in for where?

4. Dec 2, 2003

### PrudensOptimus

A, B, attract each other.

$$F_{AB} = k\frac{Q_1 Q_2}{r^2} = 6.796 * 10^{21}$$

$$E = F/q$$

Should be easy from there.

Edit: q could be thought as a differential, a number really close to 0, but never reach 0; so:

$$E = \lim_{q-->0} \frac{F}{q}$$

Last edited: Dec 2, 2003
5. Dec 6, 2003

### marshall4

How do i find the charge (q) at C for $$E = F/q$$

6. Dec 6, 2003

### marshall4

How did you figue out $$F_{AB} = k\frac{Q_1 Q_2}{r^2} = 6.796 * 10^{21}$$
k=9.0*10^9
q1= 2.5*10^-5 C
q2=-3.7*10^-7 C
r=0.35m

With those numbers i got -0.6795 N

What did i do wrong?

Also, can this problem be solved by finding electric fields separately and then adding them??

7. Dec 6, 2003

### HallsofIvy

Staff Emeritus
You've posted 5 questions all of which involve using the same basic formulas. Show us what you have done on the problem.

8. Dec 8, 2003

### marshall4

When i use the formula $$F_{AB} = k\frac{Q_1 Q_2}{r^2}$$ Do i multiply that negative sign in the equation for a negative charge?

9. Dec 8, 2003

### marshall4

This is what i did Fe=[(9.0*10^9)(2.5*10^10-5)(3.7*10^-70]/(.25)^2 =1.332

$$E = F/q$$ , but i don't know that charge at C

Is that what i do? Or what did i do wrong

PS. Is there some kind of software that i can get so i don't have to keep using ^ for exponents?

10. Dec 8, 2003

### marshall4

If i try this $$E = \lim_{q-->0} \frac{F}{q}$$ i will get

$$E = \lim_{q-->0} \frac{F(x+q)-F(x)}{q}$$ , i don't kow that the numbers are? I don't think i got the step before this

11. Dec 9, 2003

### HallsofIvy

Staff Emeritus
1. There is NO charge at C. The "electric field" (since you are treating it as a scalar) at a point is, by definition, the force that would be applied to a unit charge.

2. You don't need special software. On this forum x [ s u p]2[ / s u p] (without the spaces) will give x2.

There is a thread at the top of each forum area called
"Announcement: Howto Make Math Symbols Update" that explains that and more.

12. Dec 10, 2003

### roy5995

Would this work?

Find the electric fields a pont A & B. Then add the two fields ??? Or would you subtract the two fields ???

Use the formula $${E} = k\frac{Q}{r^2}$$

Is the field on point A & B going left or right?

Last edited: Dec 10, 2003