a charge of 4.5×10⁻⁵ c is placed in an electr...
a charge of 4.5×10⁻⁵ c is placed in an electric field with a strength of 2.0×10⁴ n/c. what is the electric force acting on the charge? n
Answer
# Explanation:
## Step1: Recall the formula
The formula for electric force is $F = qE$, where $F$ is the electric force, $q$ is the charge, and $E$ is the electric - field strength.
## Step2: Substitute the given values
Given $q = 4.5\times10^{-5}\ C$ and $E = 2.0\times10^{4}\ \frac{N}{C}$. Substitute these values into the formula: $F=(4.5\times 10^{-5}\ C)\times(2.0\times 10^{4}\ \frac{N}{C})$.
## Step3: Calculate the result
Using the rules of exponents for multiplication ($a^m\times a^n=a^{m + n}$), we have $F=(4.5\times2.0)\times10^{-5 + 4}\ N=9.0\times10^{-1}\ N = 0.9\ N$.
# Answer:
$0.9$