# Electric Field

1. Aug 26, 2005

### jena

Hi,

My Question:

What are the magnitude and direction of the electric field at a midway between a -8.0 microcolumb and a 7.0 microcolumb charge 8.0 cm apart? Assume no other charges are nearby.

Work

$$E=F/q$$

$$F=(kQ1Q2)/r^2$$

F=((9.0 X 10^9 N m^2/C)(-8.0 X10^-16 C)(7.0 X10^-6C))/(.08m)^2

F=-78.75 or 78.75 N

E=(78.75N)/(1.6X10^-19C)

E=4.92 x 10^20 N/C

Is this correct or should I look at findin the electric field using Q1 and Q2 seperately then adding them together to find a net elecrical field.

Thank you

Last edited: Aug 26, 2005
2. Aug 26, 2005

### Staff: Mentor

You need to find the field at the midpoint due to each charge, then add those fields. Don't forget: Direction matters.

Last edited: Aug 26, 2005
3. Aug 27, 2005

### lightgrav

In E=F/q , the q has to be the same *test charge* that the Force acts on!
That is, in all practical situations, the test charge q cancels.
The part of the E-field at test location t, contributed by charge Q1, is
E_1t = k Q1/(r_1t)^2 (away),
where r_1t = distance from Q1 to the test location (.04m in your case).

4. Aug 29, 2005

### jena

Oh... okay thanks