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# Electrostatics Question A spring is rigidly attached to a wall. At the end of the spring there is a charge + 15 μ C.?

A spring is rigidly attached to a wall. At the end of the spring there is a charge + 15 μ C. On a seperate wall there are three charges fixed a distance 0.7 m apart. All of the charges are equal in magnitude. The central negative charge is a distance 1.5 m from the final position of the charge attached to the spring. You can ignore any effect due to mass.

How much has the spring stretched/compressed from it's equilibrium position if the spring constant of the spring is 10 N/m

I got the spring was compressed.

### 2 Answers

- az_lenderLv 72 months agoFavorite Answer
The compressing force consists of

(2KQ^2/(1.5)^2)*(1.5)/sqrt(0.7^2 + (1.5^2)

- KQ^2/(1.5)^2

= (2KQ^2)/[(1.5)*sqrt(0.7^2 + (1.5)^2)]

- KQ^2/(1.5)^2.

Here I'm using "K" for 1/(4*pi*epsilon-nought), because I don't want to confuse it with "k" the spring constant.

The spring pushes back with a force of kx.

If I set these two forces equal, it looks like "x" is the only unknown, because K = 9.0x10^9 Nm^2/C^2 and Q = 1.5 microcoulombs, and k = 10 N/m. So now there would be a lot of button-pushing to complete the solution !!

- billrussell42Lv 72 months ago
you are missing the values of the 3 charges on the right.

"fixed a distance 0.7 m apart" is the vertical distance from the baseline to either of the + charges?

Once you supply the charge values, calculate the force on the left charge from each of the other 3, adding as vectors. The resultant will be a repulsive force, I believe (without going thru the numbers), stretching the spring.

do a first calculation on distance stretched from the above force. IF that is small compared to the 1.5 m, you are done.

IF not that means the force varies as the spring stretches, and you need calcuius to complete the problem.