Question

If a particle with a charge of +3.3 × 10?18 C is attracted to another particle by a force of 2.5 × 10?8 N, what is the magnitude of the electric field at this location? 8.3 × 10^-26 NC 1.8 × 10^10 NC 1.3 × 10^-10 N/C 7.6 × 10^9 N/C

Answer 1

7.6 × 109 N/C

Step-by-step explanation:

I got it right on the test.

Answer 2

`7.6\cdot 10^9 N/C`

Step-by-step explanation:

The relationship between force exerted on a charge and strength of the electric field is given by

`F=qE`

where

F is the strength of the electric force

q is the charge of the particle

E is the magnitude of the electric field

For the particle in the problem, we have

`q=3.3\cdot 10^(-18) C`

`F=2.5\cdot 10^(-8) N`

So the magnitude of the electric field at the location of the particle is

`E=(F)/(q)=(2.5\cdot 10^(-8))/(3.3\cdot 10^(-18))=7.6\cdot 10^9 N/C`