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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 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 $E=\sigma /\left(2 \epsilon_{0}\right)$
A charge of \(63.0 \mathrm{nC}\) is located at a distance of \(3.40 \mathrm{cm}\) from a charge of \(-47.0 \mathrm{nC} .\) What are the \(x\) - and \(y\) -components of the electric field at a point \(P\) that is directly above the 63.0 -nC charge at a distance of \(1.40 \mathrm{cm} ?\) Point \(P\) and the two charges are on the vertices of a right triangle.
An object with a charge of \(0.890 \mu \mathrm{C}\) is placed at the center of a cube. What is the electric flux through one surface of the cube?
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 \(16.8 .]\)
Suppose a charge \(q\) is placed at point \(x=0, y=0 .\) A second charge \(q\) is placed at point \(x=8.0 \mathrm{m}, y=0 .\) What charge must be placed at the point \(x=4.0 \mathrm{m}, y=0\) in order that the field at the point $x=4.0 \mathrm{m}, y=3.0 \mathrm{m} \mathrm{be}$ zero?
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