Question

3. Four charges having charge q are placed at the corners of a square with sides of length L. What is the magnitude of the force acting on any of the charges

Answer 1

Answer:

Fr = 1.91 * 9*10⁹*q²/L²

Explanation:

Let´s say that the corners of the square are  A B C and D

We are going to find out the force on the charge placed on B  ( the charge placed in the upper right corner.

As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.

Force due to charge placed in A

module   Fₓ =  K* q² / L²   in the direction of x

Force due to charge placed in C

module  Fy = K* q²/L²   in the direction of y

Force due to  the charge placed in D

That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.

d²  =  L²  +  L²  =  2*L²

d  =  √2 * L

The module of the force due to charge place in D

F₄₅ = K*q²/ 2*L²

To get the force we need to add first  Fₓ  and  Fy  

Fx + Fy  =  F₁

module of  F₁ = √ Fx² + Fy²    the direction will be the same as the diagonal of the square then:

F₁   =   √  ( K* q²/L² )²  +   ( K* q²/L² )²

F₁  =  √ 2  *  K*q²/L²

And now we add forces F₁   and F₄₅   to get the net force Fr on charge in point B.

The direction of Fr is the direction of the diagonal and is of rejection

the module is

Fr  =  F₁  *  F₄₅

Fr  =  √ 2  *  K*q²/L²  +   K*q²/ 2*L²

Fr  = ( √ 2 + 0,5 ) * K*q² /L²

K  =  9*10⁹  Nm²C²

Fr = 1.91 * 9*10⁹*q²/L²

We don´t know units of L and q