Question

Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.

SHOW YOUR WORK

Answer 1

The electric field intensity is 30000 N/C.

Step-by-step explanation:

Given:

Magnitude of the point charge is,
`q=3* 10^(-9)\ C`

Distance of the given point from the point charge is,
`d=3\ cm=0.03\ m`

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

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

Plug in
`k=9* 10^9\ N\cdot m^2/C^2, q=3* 10^(-9)\ C, d=0.03\ m`. Solve for 'E'.

`E=((9* 10^9\ N\cdot m^2/C^2)(3* 10^(-9)\ C))/((0.03\ m)^2)\\\\E=(27)/(0.0009)\ N/C\\\\E=30000\ N/C`

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.