Question

A- 5 mC charge travels due south under the influence of a 5. N force when placed into a uniform electric field. Determine the magnitude of the electric field.

Answer 1

1000 N/C

Step-by-step explanation:

The relation between electric field, electric force and charge is given as

`E=(F)/(q)`

Here E is electric field, F is the electric force and q is the electric charge.

The magnitude of electric field,

`|E|=(|F|)/(q)`

Given F = 5.0 N and q =5 mC.

Substitute the given values, we get

`|E|=(5N)/(5* 10^-^3C)`

|E| = 1000 N/C

Thus , the magnitude of the electric field is 1000 N/C.

Answer 2

Electric field will be equal to 1000 N/C

Step-by-step explanation:

We have given charge
`q=5mC=5* 10^(-3)C`

It is given that this charge is in influence of 5N

So electrostatic force F = 5 N

We have to find the electric field

Force on any charge in an electric field is equal to
`F=qE`

So
`5=5* 10^(-3)* E`

E = 1000 N/C

So electric field/ will be equal to 1000 N/C