Question

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude of the electric field at this location?

2.2 × 1026 NC
2.8 × 10−25 NC
1.5 × 1010 N/C
6.6 × 10−11 N/C

Answer 1

The magnitude of the electric field at this location is
`1.5* 10^(10) N/C`

Step-by-step explanation:

Given

`Charge\ of\ the\ particle\ Q=4.3* 10^-18\ C\\Force\ with\ which\ it\ is\ attracted\ F=6.5* 10^-8\ N`

Electric field at this location determined by the force and charge.

E=F/Q

`E=(6.5* 10^-8)/(4.3*10^-18) =1.5*\ 10^(10) \ N/C`