Question

There is a -5 nC charge present.A. Determine the magnitude and direction of the electric field at a point 7.3 meters from this charge.B. How far from this charge does the electric field have a magnitude of 2 N/C?

Answer 1

(A) E = 0.84 N/C

(B) r = 4.74 meters

Step-by-step explanation:

Given that,

Charge,
`q=-5\ nC=-5* 10^(-9)\ C`

(a) The magnitude of electric field at a point 7.3 meters from this charge is given by :

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

`E=(9* 10^9* 5* 10^(-9))/((7.3)^2)`

E = 0.84 N/C

We know that electric field due to negative charge is inwards. So, the direction of electric field is towards the charge.

(b) Let at a distance of d meters the electric field have a magnitude of 2 N/C. It is given by :

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

`r=\sqrt{(kq)/(E)}`

`r=\sqrt{(9* 10^9* 5* 10^(-9))/(2)}`

r = 4.74 meters

Hence, this is the required solution.