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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

Two metal spheres of radius \(5.0 \mathrm{cm}\) carry net charges of $+1.0 \mu \mathrm{C}\( and \)+0.2 \mu \mathrm{C} .$ (a) What (approximately) is the magnitude of the electrical repulsion on either sphere when their centers are \(1.00 \mathrm{m}\) apart? (b) Why cannot Coulomb's law be used to find the force of repulsion when their centers are \(12 \mathrm{cm}\) apart? (c) Would the actual force be larger or smaller than the result of using Coulomb's law with \(r=12 \mathrm{cm} ?\) Explain.
A hollow conducting sphere of radius \(R\) carries a negative charge \(-q .\) (a) Write expressions for the electric field \(\overrightarrow{\mathbf{E}}\) inside \((rR)\) the sphere. Also indicate the direction of the field. (b) Sketch a graph of the field strength as a function of \(r\). [Hint: See Conceptual Example \(16.8 .]\)
An electron beam in an oscilloscope is deflected by the electric field produced by oppositely charged metal plates. If the electric field between the plates is \(2.00 \times 10^{5} \mathrm{N} / \mathrm{C}\) directed downward, what is the force on each electron when it passes between the plates?
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 $E=\sigma /\left(2 \epsilon_{0}\right)$
In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
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