Question

Electric Field of a Point Charge Suppose the electric field 0.500 meters from a positive point charge is 1.25 x 105 Newtons per Coulomb. What is the charge? 0 A. 1.67 ?C (-) B. 3.48 ?C ° C. 5.14 ?C ? D. 7.29

Answer 1

Charge, q = 3.48 μC

Step-by-step explanation:

It is given that,

Electric field,
`E=1.25* 10^5\ N/C`

Distance form positive charge, r = 0.5 meters

We need to find the charge. It is given by the formula as follows :

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

`q=(Er^2)/(k)`

k = electrostatic constant

`q=(1.25* 10^5\ N/C* (0.5\ m)^2)/(9* 10^9)`

q = 0.00000347 C

or

`q=3.48* 10^(-6)\ C`

`q=3.48\ \mu C`

So, the charge is 3.48 μC. Hence, this is the required solution.