Two very small $+ 3.00 - \mu \mathrm { C }$ charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick. $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 8.99 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

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Quizplus · Accepted Answer

The electric potential due to a point charge is given by $V = k \frac{q}{r}$, where $k = 8.99 \times 10^9 \, \mathrm{N\cdot m^2/C^2}$, $q = 3.00 \mu C = 3.00 \times 10^{-6} C$, and $r = 0.5 \, \mathrm{m}$ (half the length of the meter stick to its center). Since there are two charges, the total potential at the center is the sum of the potentials due to each charge, which are equal. Thus, $V_{\text{total}} = 2 \times k \frac{q}{r} = 2 \times 8.99 \times 10^9 \times \frac{3.00 \times 10^{-6}}{0.5} = 1.08 \times 10^5 \, \mathrm{V}$.

Two Very Small +3.00−μC+ 3.00 - \mu \mathrm { C }+3.00−μC

Two Very Small $+ 3.00 - \mu \mathrm { C }$