Question

A charge Q acts on a point charge to create an electric field. Its strength, measured a distance of 40 cm away is 100 N/C. What is the magnitude of the electric field strength at a distance of 20 cm

Answer 1

`E_2=80N/C`

Step-by-step explanation:

From the question we are told that:

Initial Distance
`d_1=40cm=>0.4m`

Initial Electric field strength
`E_1=100N/C`

Final Distance
`d_2=20cm=>0.20m`

Generally the equation for Electric field is mathematically given by.

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

`q=(100*(0.4)^2)/(K)`

Substituting q for
`d=20cm`

`E_2=(k)/(0.2)*(100*(0.4)^2)/(K)`

`E_2=80N/C`