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January 31, 2015

January 31, 2015

Posted by **~christina~** on Sunday, March 30, 2008 at 12:27pm.

a) determine the magnitude and direction of the electric field at the location of charge q

b) what is the total electric force exerted on q?

basically picture is like this below

It's a square with 4 corners(circles) and one says 2q, one says q, and the bottom circles say 3q and 4q.

( I had to fill the square with .'s since I couldn't draw it without them but in reality the square is empty.

(2q)_____(q)

|.........|

|.........|

(3q)____(4q)

I don't really get how to do this.

Help please

- Physics..an electric field -
**~christina~**, Sunday, March 30, 2008 at 12:30pmBasically the drawing is of a square (a square with one particle at each corner)

Please help

- drwls=> can you help ? -
**~christina~**, Sunday, March 30, 2008 at 1:13pmplease help me if you can. with question on electric charge

- Physics..an electric field -
**drwls**, Sunday, March 30, 2008 at 1:59pmThis problem is an exercise in using Coulomb's law and adding vectors. You need to be familiar with both.

There is a force exerted by each charge and it is proportional to the product of the two charges and inversely proportional to the square of the distance between them. The force is directed along the line between the particles. You will need to know the side of the square (a) and the value of q to calculate the forces.

The force on due to 2q is directed to the right and has magnitude

F2 = 2 k q^2/a^2,

The force due to 4q has magnitude

F4 = 4 k q^2/a^2 and is directed up in the diagram.

The force due to 3q has magnitude

F3 = 3 k q^2/(2 a^2)and is directed along the diagonal from 3q to q. Note the larger denominator due to the larger separation, sqrt2 * a

k is Coulomb's constant. If you have been given no numbers for q or a, just leave it in the equations.

Your last step is to add the vector. I assume you know how to do that. Only F3 needs to be resolved into x and y components. F2 is along the x axis only and F4 is along the y axis only.

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