Two tiny grains of sand having charges of $4.0 \mu \mathrm { C }$ and $- 4.0 \mu \mathrm { C }$ are situated along the $x$-axis at $x _ { 1 } =$ $2.0 \mathrm {~m}$ and $x _ { 2 } = - 2.0 \mathrm {~m}$. What is electric potential energy of these grains relative to infinity? $( k = 1 / 4$ $\pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$ )

Question

Quizplus · Accepted Answer

The electric potential energy ($U$) between two charges is given by the formula $U = \frac{kq_1q_2}{r}$, where $k = 9.0 \times 10^9 \, \mathrm{N\cdot m^2/C^2}$, $q_1 = 4.0 \mu \mathrm{C} = 4.0 \times 10^{-6} \, \mathrm{C}$, $q_2 = -4.0 \mu \mathrm{C} = -4.0 \times 10^{-6} \, \mathrm{C}$, and $r = 4.0 \, \mathrm{m}$ (the distance between the charges). Plugging these values into the formula gives $U = \frac{(9.0 \times 10^9)(4.0 \times 10^{-6})(-4.0 \times 10^{-6})}{4.0}$, which calculates to $-36 \times 10^{-3} \, \mathrm{J}$ or $-36 \, \mathrm{mJ}$.

Two Tiny Grains of Sand Having Charges Of 4.0μC4.0 \mu \mathrm { C }4.0μC

Two Tiny Grains of Sand Having Charges Of $4.0 \mu \mathrm { C }$