Question

If a particle with a charge of +2.65 × 10−19 C is attracted to another particle by a force of 5.0 × 10−9 N, what is the magnitude of the electric field at this location?

1.9 × 10^10 N/C
5.3 × 10^−11 N/C
2.4 × 10^28 NC
1.3 × 10^−27 NC

Answer 1

Answer:

The magnitude of the electric field at this location is 1.9 × 10^10 N/C

Explanation:

In principle, the magnitude of the Electric Field ( $E$ ) is a function of the charged particle Strength ( $q$ ) and the exerted Force ( $F$ ) by:

$F=qE$      Eqn. (1).

In this problem we are given that

$q=2.65x10^{-19} C\\\\F=5.0x10^{-9}N$

So we can simply rearrange Eqn. (1), plug in the given values and solve for $E$ as follow:

$E=\frac{F}{q}\\ \\E=\frac{5.0x10^{-9}}{2.65x10^{-19}}\\ \\E=1.886x10^{10}\\\\E=1.9x10^{10}N/C$

Therefore, Option A) is correct from the available options.

Answer 2

Answer:

Your answer would be the first one :), 1.9 × 10^10 N/C hope this helped.

Explanation: