Two charges, each of 2.9 microc are placed at two corners of a square 50cm on a side, if the charges are on one side of the square what would be the magnitude and direction of the net electric field a one of the other corners of the square?

The magnitude of the electric field and direction of electric field are and 75.36°.

Explanation:

Given that,

First charge

Second charge

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

Put the value into the formula

For E₂,

Using formula of electric field

Put the value into the formula

We need to calculate the horizontal electric field

We need to calculate the vertical electric field

We need to calculate the net electric field

Put the value into the formula

We need to calculate the direction of electric field

Using formula of direction

Hence, The magnitude of the electric field and direction of electric field are and 75.36°.

insects!

explanation:

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.