Three equal charges of magnitude 'e' are located at the vertices of an equilateral triangle of side 1m. where should you place a charge of 2e so that the 2e' does not experience any force?

We would have to locate the charge at the center of the triangle.

Explanation:

I tried my best to make geometric arguments, you can see the attached picture if it helps.

Let's say the triangle has three corners names 1, 2 and 3 as seen in the image. Let's determine the electric field at the center of the triangle. We can see that the x-component of electric field produced by the charge at 1 and the one produce by the charge at 2 will cancel each other out because they have the same but opposite values. The y components on the other hand are equal and are added. The magnitude of all three electric fields are the same (because each corner is at equal distance from the center and all three charges are equal, see the formula for the electric field: where is the distance between the point of observation and the charge). Thus lets call E the magnitude of the electric field produced by any of the charges.

We can clearly see that the y-component of the electric filed produced by 1 is , the same goes for the one produced by the charge at 2, knowing that , we have that the total field due to the charges at 1 and 2 are equal to E in the positive y-direction. The field produced by the third charge is clearly -E in the negative y-direction thus canceling the contribution from 1 and 2. The total electric field at the center of the triangle is zero, this also mean that any charge placed at that point will not feel any force.

the answer is kevlar

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.