Question

What is the magnitude of the electric field 17.1 cm directly above an isolated 1.83Ã10â5 C charge?

Answer 1

Electric field,
`E=5.63* 10^(16)\ N/C`

Step-by-step explanation:

Given that,

Charge,
`q=1.83* 10^5\ C`

We need to find the magnitude of electric field 17.1 cm (0.171 m) above an isolated charge. Electric field at a point is given by :

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

`E=(9* 10^9* 1.83* 10^5\ C)/((0.171\ m)^2)`

`E=5.63* 10^(16)\ N/C`

So, the electric field is
`5.63* 10^(16)\ N/C`. Hence, this is the required solution.