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7. The three charges in the figure below are at the vertices of an isosceles triangle. Let q = 7.50 nC and calculate the elec
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(7) The electric potential is

V=V_{1}+V_{2}+V_{3}

V=\frac{kq_{1}}{r_{1}}+\frac{kq_{2}}{r_{2}}+\frac{kq_{3}}{r_{3}}

V=-\frac{kq}{d_{1}/2}-\frac{kq}{d_{1}/2}+\frac{kq}{\sqrt{d_{2}^{2}-(\frac{d_{1}}{2})^{2}}}

V=-\frac{4kq}{d_{1}}+\frac{kq}{\sqrt{d_{2}^{2}-(\frac{d_{1}}{2})^{2}}}

V=-\frac{4*9.*10^{9}*7.50*10^{-9}}{0.01}+\frac{9.*10^{9}*7.50*10^{-9}}{\sqrt{0.07^{2}-(\frac{0.01}{2})^{2}}}

V=-27000+966.76

V=-26033V

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