# Coulomb's Law Problem magnitude and direction

swede5670

## Homework Statement

What are the magnitude and direction of the electric field at a point midway between a -9.0 µC and a +6.0 µC charge 3.0 cm apart? Assume no other charges are nearby.

## Homework Equations

Fe= {K(q1)(q2)}/d^2

k = (8.99 x 10^9)

## The Attempt at a Solution

First I converted the µC to C =
-9 µC = -.000009 C
6 µC = .000006

Then convert 3 cm to mm = .03

(8.99 x 10^9) (-.000009) ( .000006 ) = -.48546

-.48546/d^2

-.48546/ .03

-539.4

My final answer is -539.4 but I know its wrong, can anyone tell me what I did incorrectly?

Homework Helper

## Homework Statement

What are the magnitude and direction of the electric field at a point midway between a -9.0 µC and a +6.0 µC charge 3.0 cm apart? Assume no other charges are nearby.

## Homework Equations

Fe= {K(q1)(q2)}/d^2
k = (8.99 x 10^9)

## The Attempt at a Solution

First I converted the µC to C =
-9 µC = -.000009 C
6 µC = .000006

Then convert 3 cm to mm = .03

(8.99 x 10^9) (-.000009) ( .000006 ) = -.48546
-.48546/d^2
-.48546/ .03
-539.4

My final answer is -539.4 but I know its wrong, can anyone tell me what I did incorrectly?

You found the Force between the 2 charges. The question asks for the field at a point half way between the two.

F = q*E

What you want is E, given by

E = kq/r2

Your distance is .015 m to each. And remember these are vectors and when you add the field at the mid point you want to take care with what direction the field of each would be pointing.

swede5670
What Q value would you use though?

Homework Helper
What Q value would you use though?

For E you are using the q of the charges you are measuring. 1 term will be the field from the 6 µC charge and the other the 9 µC.

swede5670
So there will therefore be two answers?

I'm confused by what your saying
"For E you are using the q of the charges you are measuring."
I realize this, but do I need to set up two equations then? Or do I have to add the charges?

Homework Helper
So there will therefore be two answers?

I'm confused by what your saying
"For E you are using the q of the charges you are measuring."
I realize this, but do I need to set up two equations then? Or do I have to add the charges?

The E-field at any point will be the ∑ of the E from all charges

swede5670
so sigma is the sum right?

E = k(6x10^-6)/r2
E = k(-9*10^-6)/r2

once I've added these two Es up is that my final answer?

Homework Helper
so sigma is the sum right?

E = k(6x10^-6)/r2
E = k(-9*10^-6)/r2

once I've added these two Es up is that my final answer?

Sort of. But you must be careful with the sign. Since one charge is plus and the other minus, in between the charges then the field magnitudes both add and point toward the negative charge.

swede5670
{(8.99*10^9)(-9*10^-6)}/.015^2 = -359,600,000

{(8.99*10^9)(6*10^-6)}/.015^2 = 239, 733, 333

-359,600,000 + 239, 733, 333 = -119, 866, 666

-119, 866, 666 N/C seems ridiculous, what did I do here?

Homework Helper
{(8.99*10^9)(-9*10^-6)}/.015^2 = -359,600,000

{(8.99*10^9)(6*10^-6)}/.015^2 = 239, 733, 333

-359,600,000 + 239, 733, 333 = -119, 866, 666

-119, 866, 666 N/C seems ridiculous, what did I do here?

Not quite.

As I said several times, these are vectors. As it turns out both vectors are pointing in the same direction.

To determine the |F| then you simply add the |Fq1| + |Fq2|.

The - sign means that the Force points toward a charge and the + sign means that the field points away from the charge. Draw a picture nd you will see that at the mid point for these charges the E-field from both is pointing in the same direction ... hence the magnitudes add at that point.