Question

during a lightning discharge, 35.8 c of charge moves through a potential difference of 2.98 × 10⁸ v in 1.73 × 10⁻² s. the energy released by this lightning bolt is j.

Explanation:

Step1: Recall the formula for electric potential energy

The formula for the energy \( E \) released when a charge \( q \) moves through a potential difference \( V \) is \( E = qV \).

Step2: Identify the given values

We are given \( q = 35.8 \, \text{C} \) and \( V = 2.98\times 10^{8} \, \text{V} \).

Step3: Substitute the values into the formula

\[

$$\begin{align*} E&=qV\\ &=35.8\times(2.98\times 10^{8})\\ &=35.8\times2.98\times 10^{8}\\ & = 106.684\times 10^{8}\\ &=1.06684\times 10^{10} \end{align*}$$

\]

Answer:

\( 1.07\times 10^{10} \) (rounded to three significant figures) or \( 1.06684\times 10^{10} \) (exact value from calculation)