Answer to Question #127172 in Electric Circuits for Alexandra G Castellanos

Question

A charge of -2.790 μC is located at (2.925 m , 4.726 m ), and a charge of 1.460 μC is located at (-2.696 m , 0). There is one point on the line connecting these two charges where the potential is zero. Find this point. x,y= m

Answer 1

The potential of an point charge can be calculated as $\varphi = k\cdot\dfrac{q}{r}.$ Let (x,y) be the coordinates of the point in question, (x₁,y₁) and (x₂,y₂) be the coordinates of the charges. Therefore we should solve an equation

$k\cdot\dfrac{Q_1}{\sqrt{(x_1-x)^2+(y_1-y)^2}} + k\cdot\dfrac{Q_2}{\sqrt{(x_2-x)^2+(y_2-y)^2}} = 0.$ Or

$\dfrac{Q_1^2}{{(x_1-x)^2+(y_1-y)^2}} = \dfrac{Q_2^2}{{(x_2-x)^2+(y_2-y)^2}}$ .

$Q_1^2(x_2-x)^2+Q_1^2(y_2-y)^2 = Q_2^2(x_1-x)^2+Q_2^2(y_1-y)^2.$

We should find a point on the line connecting two points, so $y = y_1+(x-x_1)\cdot\dfrac{y_2-y_1}{x_2-x_1}.$

Next, we substitute this value into the last equation and solve it with respect to x. We get two values of x and choose the value that is situated between two points with charges.

The solutions of the equation are x=-0.765 m and x=-8.866 m, so we choose x=-0.765 m. Next we calculate y, so it is 1.62 m.