Electric field and electric force

[-EquinoX-]

Homework Statement

What is the electric field at location of charge q

Homework Equations

The Attempt at a Solution

My intuition in solving this is by finding the field of each charges relative to q and add them all, so for example for the 2q charges relative to q, the electric forces it contributes is ke * 2q/a^2, am I right?

 

[rl.bhat]

You have to find the magnitude and direction of forces due to three charges at q, and then find the vector sum of all forces.

 

[-EquinoX-]

That's what I was trying to say, so say the magnitude of the force 2q at q is ke * 2q/a^2 in the positive x direction (or i hat) , am I right?

 

[rl.bhat]

Yes. You right. Similarly find the forces due to 3q and 4q, and find the vector sum.

 

[-EquinoX-]

hmm..for the 3q charge, the distance to q is sqrt(2)a, so therefore the magnitude can be found by:

ke * 3q/(sqrt(2)a)^2 cos(45) i + ke * 3q/(sqrt(2)a)^2 sin(45) j

is this correct?

 

[rl.bhat]

hmm..for the 3q charge, the distance to q is sqrt(2)a, so therefore the magnitude can be found by:

ke * 3q/(sqrt(2)a)^2 cos(45) i + ke * 3q/(sqrt(2)a)^2 sin(45) j

is this correct?

You can directly wright down as ke*3q/2a^2 for 3q and ke*4q/a^2.
Field due 2q and 4q are perpendicular to each other. Find the resultant of these fields. Add it to the field due to 3q to get the final field.

 

[-EquinoX-]

>>Field due 2q and 4q are perpendicular to each other. Find the resultant of these fields.

Well I already did find the field due 2q and q right? Why do I have to find between 2q and 41? The question asks with respect to q

 

[rl.bhat]

The field at location of charge q is the vector sum of fields due to 2q, 3q and 4q.

 

[-EquinoX-]

Thats trueand so why is my answer 2 posts above is wrong?