Question

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 \(A ?\)

Short answer

Now that we have determined the position of charge Q3, write a short answer discussing its significance and how it affects the electric field at point A. By placing the third charge Q3 at approximately 0.212 m from the midpoint of the line connecting charges Q1 and Q2, the electric field at point A becomes zero. This position is significant because it ensures that the electric field created by Q3 cancels out the combined electric field created by Q1 and Q2 at point A. As a result, there is no net electric field at point A, which could be beneficial in certain applications where maintaining a neutral or balanced electric field is important.

Step-by-step solution

Introducing the problem and variables

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.

Calculate the electric field at point A due to charges \(Q_1\) and \(Q_2\)

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: $$\textbf{E}_{1A} + \textbf{E}_{2A} = \textbf{E}_{12A}$$

Calculate the electric field at point A due to charge \(Q_3\)

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, \(|\textbf{E}_{3A}| = |\textbf{E}_{12A}|\).

Calculate the position of charge \(Q_3\)

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 \(Q_3\).

Solve for the distance \(r_3\)

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 \(Q_2\).