Question

A 90 μC point charge is at the origin.

(a) Find the electric field at the point x1= 45 cm , y1= 0
(b) Find the electric field at the point x3= -20 cm , y3 = 60 cm.

Answer 1

a)
`E_1=4* 10^6\ N.C^(-1)`

b)
`E_3=202.5\ N.C^(-1)`

Step-by-step explanation:

Given:

charge at the origin,
`q=90* 10^(-6)\ C`

a)

Electric field at point
`(45\ cm,0\ cm)`:

The distance form the charge:

`d_1=0.45\ m`

Now the electric field at the given point:

`E_1=(1)/(4\pi.\epsilon_0) * (q)/(d_1^2)`

`E_1=9* 10^9* (90* 10^(-6))/(0.45^2)`

`E_1=4* 10^6\ N.C^(-1)`

b)

Electric field at point
`(-20\ cm, 60\ cm)`:

The distance form the charge:

`d_3=√((-20-0)^2+(60-0)^2)`

`d_3=63.2456\ m`

Now the electric field at the given point:

`E_3=(1)/(4\pi.\epsilon_0) * (q)/(d_3^2)`

`E_3=9* 10^9* (90* 10^(-6))/(4000)`

`E_3=202.5\ N.C^(-1)`