Question

at the vertices of a square with a side of 5 cm, there are identical positive charges q = 2 nC. determine the strength of the electrostatic field in the middle of one of the sides of the square

Answer 1

Step-by-step explanation:

To make this problem the easiest way possible, draw a picture and choose the side between the charges. The field will be zero at that point, and I'll prove it in just a second.

`E=(kq)/(r^2)`

k is a constant with a value of
`8.99*10^9Nm^2/C^2`

q is the magnitude of the charge producing the field

r is the distance from the source charge to the test charge

So first determine the electric field from the charge on the bottom left corner, then we'll determine the electric from the bottom right corner.

Convert the centimeters to meters and nano-Coulombs to Coulombs

5cm = 0.05m

2 nC = 2 x 10^-9 C

`E=(8.99*10^9)(2*10^-^9)/(0.025)^2=2.8768*10^4N/C`

This is pointing to the right because electric field lines point away from positive charges.

`E=(8.99*10^9)(2*10^-^9)/(0.025)^2=-2.8768*10^4N/C`

This is pointing to the left because of the same reason. Field lines point away from positive charges.

You are able to sum them up because they are both in the x-direction. Their sum will be a net field value of zero.