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Chapter 16: Problem 40

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

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

Step by step solution

01

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.
02

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: \(r = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2} = \frac{a}{\sqrt{2}}\)
03

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: \(E_1 \approx E_2 \approx 1.18 \times 10^5 \ \frac{N}{C}\)
04

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: \(E_{total} \approx 1.67 \times 10^5 \ \frac{N}{C}\)

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Most popular questions from this chapter

A point charge \(q_{1}=+5.0 \mu \mathrm{C}\) is fixed in place at \(x=0\) and a point charge \(q_{2}=-3.0 \mu \mathrm{C}\) is fixed at \(x=\) $-20.0 \mathrm{cm} .\( Where can we place a point charge \)q_{3}=-8.0 \mu \mathrm{C}$ so that the net electric force on \(q_{1}\) due to \(q_{2}\) and \(q_{3}\) is zero?
\(\mathrm{A}+2.0\) -nC point charge is \(3.0 \mathrm{cm}\) away from a $-3.0 \mathrm{-nC}$ point charge. (a) What are the magnitude and direction of the electric force acting on the +2.0 -nC charge? (b) What are the magnitude and direction of the electric force acting on the -3.0 -nC charge?
Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?
Three equal charges are placed on three corners of a square. If the force that \(Q_{\mathrm{a}}\) exerts on \(Q_{\mathrm{b}}\) has magnitude \(F_{\mathrm{ba}}\) and the force that \(Q_{\mathrm{a}}\) exerts on \(Q_{\mathrm{c}}\) has magnitude \(F_{\mathrm{ca}},\) what is the ratio of \(F_{\mathrm{ca}}\) to $F_{\mathrm{ba}} ?$
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
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