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. Find the electric field at point \(C\) the center of the square.

Short answer

Answer: The total electric field at the center of the square is approximately 1.67 x 10^5 N/C.

Step-by-step solution

Identify the given values and formula to use

We are given the following values: Charge on both objects: \(q = 7.00 \ \mu C = 7.00 \times 10^{-6} \ C\) Side of the square: \(a = 0.300 \ m\) We will be using the formula for the electric field due to a point charge \(E = \frac{k \times q}{r^2}\), where \(k = 9 \times 10^{9} \ \frac{Nm^2}{C^2}\) is the electrostatic constant.

Calculate the distance of each charged object from point C

In the given square, we can observe that the two charged objects are symmetrically located with respect to point C. Each charged object is located at distance equal to half of the diagonal of the square from the center of the square (point C). Using the Pythagorean theorem to find the distance: \(r = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2} = \frac{a}{\sqrt{2}}\)

Calculate the electric field due to each charged object at point C

Using the point charge formula we derived earlier: \(E_1 = E_2 = \frac{k \times q}{r^2} = \frac{9 \times 10^9 \times 7.00 \times 10^{-6}}{(\frac{0.300}{\sqrt{2}})^2} \ \frac{N}{C}\) Calculating the value of the electric field due to each charge: \(E_1 \approx E_2 \approx 1.18 \times 10^5 \ \frac{N}{C}\)

Calculate the total electric field at point C

The total electric field at point C is the vector sum of the electric fields due to both charged objects. Since the charged objects are symmetric with respect to point C, their electric fields form a right triangle at point C. Using the Pythagorean theorem to find the magnitude of the resultant electric field: \(E_{total} = \sqrt{ E_1^2 + E_2^2 } = \sqrt{(1.18 \times 10^5)^2 + (1.18 \times 10^5)^2} = 1.18 \times 10^5 \cdot \sqrt{2} \ \frac{N}{C}\) The total electric field at point C is approximately: \(E_{total} \approx 1.67 \times 10^5 \ \frac{N}{C}\)