Question

\(\mathrm{~A}+1.60-\mathrm{nC}\) point charge is placed at one corner of a square \((1.00 \mathrm{~m}\) on a side), and a \(-2.40-\mathrm{nC}\) charge is placed on the corner diagonally opposite. What is the magnitude of the electric field at either of the other two corners?

Short answer

Short Answer: The magnitude of the electric field at either of the other two corners is \(2.256 \times 10^4 \cdot \sqrt{(A+1.60\times10^{-9})^2 + (2.40\times10^{-9})^2} \frac{N}{C}\), where A is the unknown charge.

Step-by-step solution

Identify the System and Relevant Information

We have a square with side length 1.00 m. In one corner, we have a charge of (A + 1.60 nC), and the diagonally opposite corner holds a charge of -2.40 nC. We need to find the magnitude of the electric field at either of the other two corners.

Calculate the Distance Between the Charges and the Given Corners

Since the charges are placed diagonally, we can calculate the distance using the Pythagorean theorem. Let's call this distance d. We have: \(d = \sqrt{(1.00)^2 + (1.00)^2} = \sqrt{2} \mathrm{~m}\)

Find the Magnitude of the Electric Field Due to Each Charge

The magnitude of the electric field due to a point charge is given by: \(E = \frac{k|q|}{r^2}\) Here, k is the electrostatic constant, approximately equal to \(8.99\times10^9 \frac{N \cdot m^2}{C^2}\), q is the charge, and r is the distance from the charge. Now let's find the electric fields due to the given charges: Charge 1: \((A+1.60) \times 10^{-9} C\) \(E_1 = \frac{8.99\times10^9 \frac{N \cdot m^2}{C^2} \times|(A+1.60) \times 10^{-9} C|}{(\sqrt{2} \mathrm{~m})^2}\) Charge 2: \((-2.40) \times 10^{-9} C\) \(E_2 = \frac{8.99\times10^9 \frac{N \cdot m^2}{C^2} \times |-2.40 \times 10^{-9} C|}{(\sqrt{2} \mathrm{~m})^2}\)

Calculate the Vector Sum of Electric Fields

The superposition principle states that the total electric field at a given point is the vector sum of the electric fields due to each charge. Here, since we are considering the square's corners, the electric fields due to the charges are at right angles to each other. As a result, we can use the Pythagorean theorem to find the magnitude of the resultant electric field. \(E_\mathrm{resultant} = \sqrt{E_1^2 + E_2^2}\)

Substitute the Known Values and Simplify

Now, substitute the expressions for \(E_1\) and \(E_2\) calculated in step 3, and simplify: \(E_\mathrm{resultant} = \sqrt{\left(\frac{8.99\times10^9 (A+1.60 \times 10^{-9})}{2}\right)^2 + \left(\frac{8.99\times10^9 (2.40 \times 10^{-9})}{2}\right)^2}\)

Calculate the Magnitude of the Resultant Electric Field

Finally, calculate the magnitude of the electric field: \(E_\mathrm{resultant} = 2.256 \times 10^4 \cdot \sqrt{(A+1.60\times10^{-9})^2 + (2.40\times10^{-9})^2} \frac{N}{C}\) The magnitude of the electric field at either of the other two corners of the square is \(2.256 \times 10^4 \cdot \sqrt{(A+1.60\times10^{-9})^2 + (2.40\times10^{-9})^2} \frac{N}{C}\).