Three equal charges 1.8 10^ 6\ C each are located at the corners of an equilateral triangle ABC of side 10\ cm . Calculate the electric potential due to these charges at the mid point of AB.

Solution

Given:

• Three equal charges, each $q=1.8\times 10^{-6}\mathrm{C}$, placed at the corners of an equilateral triangle $ABC$ with side length $10\mathrm{cm}=0.1\mathrm{m}$.
• We need to calculate the electric potential at the midpoint $M$ of side $AB$.

Step 1: Understand the geometry

• The midpoint $M$ of $AB$ is equidistant from $A$ and $B$, so the distance from $M$ to $A$ or $B$ is: $$r_{AM}=r_{BM}=\frac{10}{2}=5\mathrm{cm}=0.05\mathrm{m}.$$
• The distance from $M$ to $C$ can be calculated using the Pythagorean theorem: $$r_{CM}=\sqrt{{\left(\frac{10}{2}\right)}^2+{\left(\frac{10\sqrt{3}}{2}\right)}^2}=\sqrt{25+75}=\sqrt{100}=10\mathrm{cm}=0.1\mathrm{m}.$$

Step 2: Recall the formula for electric potential The electric potential $V$ at a point due to a charge $q$ at a distance $r$ is given by: $$V=\frac{kq}{r},$$ where $k=9\times 10^9\mathrm{N}\cdot {\mathrm{m}}^2\cdot {\mathrm{C}}^{-2}$ is Coulomb’s constant.

Step 3: Calculate the potential due to each charge

• Potential due to charge at $A$: $$V_A=\frac{kq}{r_{AM}}=\frac{9\times 10^9\times 1.8\times 10^{-6}}{0.05}.$$
• Potential due to charge at $B$: $$V_B=\frac{kq}{r_{BM}}=\frac{9\times 10^9\times 1.8\times 10^{-6}}{0.05}.$$
• Potential due to charge at $C$: $$V_C=\frac{kq}{r_{CM}}=\frac{9\times 10^9\times 1.8\times 10^{-6}}{0.1}.$$

Step 4: Compute numerical values $$V_A=V_B=\frac{9\times 10^9\times 1.8\times 10^{-6}}{0.05}=3.24\times 10^5\mathrm{V},$$ $$V_C=\frac{9\times 10^9\times 1.8\times 10^{-6}}{0.1}=1.62\times 10^5\mathrm{V}.$$

Step 5: Total electric potential at $M$ Since electric potential is a scalar quantity, the total potential at $M$ is the algebraic sum of the potentials due to all three charges: $$V_{total}=V_A+V_B+V_C=3.24\times 10^5+3.24\times 10^5+1.62\times 10^5=8.1\times 10^5\mathrm{V}.$$

Final Answer: The electric potential at the midpoint of $AB$ is $8.1\times 10^5\mathrm{V}$.