Question

a charged particle is located in an electric field where the magnitude of the electric field strength is 2.0×10³ n/c. if the electrostatic force exerted on the particle is 3.0×10⁻³ n, what is the charge of the particle? a 1.5×10⁻⁶ c b 6.0 c

Explanation:

Step1: Recall the formula for electric force

The formula for the electrostatic force \(F\) on a charged particle in an electric field \(E\) is \(F = qE\), where \(q\) is the charge of the particle.

Step2: Solve for the charge \(q\)

Rearrange the formula to \(q=\frac{F}{E}\).
Substitute \(F = 3.0\times10^{-3}\space N\) and \(E=2.0\times 10^{3}\space N/C\) into the formula:
\(q=\frac{3.0\times 10^{-3}}{2.0\times 10^{3}}\)
Using the rule of exponents \(\frac{a^{m}}{a^{n}}=a^{m - n}\), we have \(q=\frac{3.0}{2.0}\times10^{-3-3}\)
\(q = 1.5\times10^{-6}\space C\)

Answer:

A. \(1.5\times 10^{-6}\space C\)