Two $+ 6.0 - \mu \mathrm { C }$ charges are placed at two of the vertices of an equilateral triangle having sides $2.0 \mathrm {~m}$ Iong. What is the electric potential at the third vertex, relative to infinity, due to these charges? $( k =$ $1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$ )

Question

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The electric potential at a point due to a point charge is given by $V = k \frac{q}{r}$, where $k$ is Coulomb's constant, $q$ is the charge, and $r$ is the distance from the charge to the point. Since there are two $+6.0 \mu C$ charges and the distance from each charge to the third vertex is $2.0 m$, the total potential at the third vertex is the sum of the potentials due to each charge: $V_{total} = 2 \times k \frac{6.0 \times 10^{-6} C}{2.0 m} = 9.0 \times 10^9 \frac{N \cdot m^2}{C^2} \times 3.0 \times 10^{-6} C = 27 kV$.

Two +6.0−μC+ 6.0 - \mu \mathrm { C }+6.0−μC Charges Are Placed at Two of the Vertices of an Are

Two $+ 6.0 - \mu \mathrm { C }$ Charges Are Placed at Two of the Vertices of an Are