Question

What magnitude charge creates a 32.08 N/C electric field at a point 2.84 m away?

Answer 1

Q = 29.4 x 10⁻⁹ C.

Step-by-step explanation:

Electric field due to a charge Q at distance d is given by coulomb law as follows

Electric field

E = k Q /d²

where for air k which is a constant is 9 x 10⁹

`E=(k* Q)/(d^2)`

Given E = 32.08 , d = 2.84 m

Putting these values in the relation above, we have

`[tex]32.08=(9* 10^9* Q)/((2.84)^2)`[/tex]

Q = 29.4 x 10⁻⁹ C.