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 $1.5\times 10^{10} N/C$

Explanation:

Given

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

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

E=F/Q

$E=\frac{6.5\times 10^-8}{4.3\times10^-18} =1.5\times\ 10^{10} \ N/C$