calculating electric potential energy\na charge of 4.5×10⁻⁵ c is placed in an electric field with a strength…

calculating electric potential energy\na charge of 4.5×10⁻⁵ c is placed in an electric field with a strength of 2.0×10⁴ n/c. if the charge is 0.030 m from the source of the electric field, what is the electric potential energy of the charge?\nj

calculating electric potential energy\na charge of 4.5×10⁻⁵ c is placed in an electric field with a strength of 2.0×10⁴ n/c. if the charge is 0.030 m from the source of the electric field, what is the electric potential energy of the charge?\nj

Answer

Explanation:

Step1: Recall the formula for electric potential energy

The formula for electric - potential energy $U = qEd$, where $q$ is the charge, $E$ is the electric - field strength, and $d$ is the distance from the source of the electric field.

Step2: Identify the given values

$q = 4.5\times10^{-5}\text{ C}$, $E = 2.0\times10^{4}\frac{\text{N}}{\text{C}}$, and $d = 0.030\text{ m}$.

Step3: Substitute the values into the formula

$U=(4.5\times 10^{-5}\text{ C})\times(2.0\times 10^{4}\frac{\text{N}}{\text{C}})\times(0.030\text{ m})$. First, multiply the numerical values: $4.5\times2.0\times0.030 = 4.5\times0.060=0.27$. Then, multiply the powers of 10: $10^{-5}\times10^{4}=10^{-5 + 4}=10^{-1}$. So, $U = 0.27\times10^{-1}\text{ J}=0.027\text{ J}$.

Answer:

$0.027$