Electric Field at Point C: A to B & B to C

[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.

 

[PrudensOptimus]

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

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

 

[marshall4]

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

 

[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:

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

 

[marshall4]

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

 

[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??

 

[HallsofIvy]

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

 

[marshall4]

Originally posted by 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??

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?

 

[marshall4]

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

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?

 

[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

 

[HallsofIvy]

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.

 

[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?