Two tiny objects with equal charges of \(7.00 \mu \mathrm{C}\) are placed at the two lower corners of a square with sides of \(0.300 \mathrm{m},\) as shown. Where would you place a third small object with the same charge so that the electric field is zero at the corner of the square labeled \(A ?\)

Let's label the lower left charge as \(Q_1\), the lower right charge as \(Q_2\), and the third charge that we want to place as \(Q_3\). The distance between \(Q_1\) and \(Q_2\) is \(0.3 m\), and \(Q_1\), \(Q_2\), and \(Q_3\) all have a charge of \(7.00 \mu C\). The distances from \(Q_1\), \(Q_2\), and \(Q_3\) to point A are \(r_1\), \(r_2\), and \(r_3\), respectively.

Using Coulomb's law, the electric field at point A due to charge \(Q_1\) is: $$\textbf{E}_{1A} = \frac{k Q_1}{r_1^2}$$ and the electric field at point A due to charge \(Q_2\) is: $$\textbf{E}_{2A} = \frac{k Q_2}{r_2^2}$$ Since all charges are equal, both electric fields have the same magnitude: $$\left|\textbf{E}_{1A}\right| = \left|\textbf{E}_{2A}\right|$$ Since \(Q_1\) and \(Q_2\) are on a straight line, the two electric fields at point A add up: $$\textbf{E}_{1A} + \textbf{E}_{2A} = \textbf{E}_{12A}$$

Using Coulomb's law again, the electric field at point A due to charge \(Q_3\) is: $$\textbf{E}_{3A} = \frac{k Q_3}{r_3^2}$$ To make the total electric field at point A zero, \(\textbf{E}_{3A}\) must cancel out \(\textbf{E}_{12A}\). Therefore, \(|\textbf{E}_{3A}| = |\textbf{E}_{12A}|\).

Since we have the following equation for the magnitudes of the electric fields at point A: $$|\textbf{E}_{1A}| + |\textbf{E}_{2A}| = |\textbf{E}_{3A}|$$ We plug in the expressions of each electric field in terms of their respective distances from point A: $$\frac{k Q_1}{r_1^2} + \frac{k Q_2}{r_2^2} = \frac{k Q_3}{r_3^2}$$ From problem statement, we know that \(Q_1 = Q_2 = Q_3 = 7\mu C\). Thus, we can simplify the equation by cancelling the charge terms and Coulomb constant: $$\frac{1}{r_1^2} + \frac{1}{r_2^2} = \frac{1}{r_3^2}$$ We know that \(r_1 = r_2 = 0.3 m\), and \(r_3\) is the distance from the midpoint of the line connecting \(Q_1\) and \(Q_2\) to point A: $$\frac{1}{(0.3)^2} + \frac{1}{(0.3)^2} = \frac{1}{r_3^2}$$ Solve for \(r_3\) to get the position of charge \(Q_3\).

Solving the equation: $$\frac{1}{(0.3)^2} + \frac{1}{(0.3)^2} = \frac{1}{r_3^2}$$ We find: $$r_3 = \frac{1}{\sqrt{2(0.3)^{-2}}} = \frac{1}{\sqrt{2(3.33)^{2}}} \approx 0.212 m$$ So, to make the electric field at point A zero, we should place the third charge \(Q_3\) at approximately \(0.212 m\) from the midpoint of the line connecting charges \(Q_1\) and \(Q_2\).