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

Answer:

$7.6\cdot 10^9 N/C$

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

$E=\frac{F}{q}=\frac{2.5\cdot 10^{-8}}{3.3\cdot 10^{-18}}=7.6\cdot 10^9 N/C$