The electric potential at a distance of 2.0 * 10^-10 m from a nucleus of charge +12e is

Question

Quizplus · Accepted Answer

The electric potential $V$ due to a point charge (such as a nucleus) at a distance $r$ is given by the formula $V = \frac{kQ}{r}$, where $k$ is Coulomb's constant ($8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$), $Q$ is the charge, and $r$ is the distance from the charge. For a nucleus with a charge of +12e, where $e = 1.6 \times 10^{-19} \, \text{C}$ is the elementary charge, the charge $Q = 12e = 12 \times 1.6 \times 10^{-19} \, \text{C}$. Given that $r = 2.0 \times 10^{-10} \, \text{m}$, we can calculate the potential as follows:$$V = \frac{(8.99 \times 10^9) \times (12 \times 1.6 \times 10^{-19})}{2.0 \times 10^{-10}} = \frac{(8.99 \times 10^9) \times (19.2 \times 10^{-19})}{2.0 \times 10^{-10}} = 86 \, \text{V}.$$Therefore, the correct answer is 86 V.

The Electric Potential at a Distance of 2