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Chapter 16: Problem 64

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?

Short Answer

Expert verified
Answer: The electric flux through one surface of the cube is 1.662 x 10鲁 N.m虏/C.

Step by step solution

01

Understanding Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is given by the formula: 鈭瓻 鈰� dA = Q/蔚鈧� Where E is the electric field, dA is a surface area element, Q is the enclosed charge, and 蔚鈧� is the vacuum permittivity (蔚鈧� = 8.85 x 10鈦宦孤� C虏/N.m虏).
02

Electric field and the cube

In our problem, there is a charge placed at the center of the cube. The electric field produced by the charge is radially outward and symmetric in all directions. So, the electric field would be same through all faces of the cube.
03

Use Gauss's law to find flux through the entire cube

For the entire cube, we can write Gauss's Law as: 桅 = 鈭瓻 鈰� dA = Q/蔚鈧� Here, the electric field E has the same value on each face of the cube, so we can rewrite this as: 桅 = E * (area of all faces) = Q/蔚鈧�
04

Calculate the area of all faces

We know that there are 6 faces in the cube and the electric field on each face is the same. Let's denote each face's area as A. So, the area of all faces will be: (area of all faces) = 6A 桅 = E * 6A = Q/蔚鈧�
05

Find the electric flux through one surface of the cube

We want to find the electric flux through only one face of the cube (A). We can divide both sides of the equation by 6: (E * 6A)/6 = (Q/蔚鈧�)/6 E * A = (Q/6蔚鈧�) Now, we have the electric flux through one face of the cube: 桅鈧� = (Q/6蔚鈧�)
06

Plug in values

We know the given charge Q = 0.890 渭C = 0.890 x 10鈦烩伓 C, and 蔚鈧� = 8.85 x 10鈦宦孤� C虏/N.m虏. Plug in these values: 桅鈧� = ((0.890 x 10鈦烩伓)/(6 * 8.85 x 10鈦宦孤�)) 桅鈧� = 1.662 x 10鲁 N.m虏/C So, the electric flux through one surface of the cube is 1.662 x 10鲁 N.m虏/C.

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Most popular questions from this chapter

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 \(16.8 .]\)
A raindrop inside a thundercloud has charge - \(8 e\). What is the electric force on the raindrop if the electric field at its location (due to other charges in the cloud) has magnitude $2.0 \times 10^{6} \mathrm{N} / \mathrm{C}$ and is directed upward?

Two point charges are separated by a distance \(r\) and repel each other with a force \(F\). If their separation is reduced to 0.25 times the original value, what is the magnitude of the force of repulsion between them?
(a) Use Gauss's law to prove that the electric field outside any spherically symmetric charge distribution is the same as if all of the charge were concentrated into a point charge. (b) Now use Gauss's law to prove that the electric field inside a spherically symmetric charge distribution is zero if none of the charge is at a distance from the center less than that of the point where we determine the field.
The electric field across a cellular membrane is $1.0 \times 10^{7} \mathrm{N} / \mathrm{C}$ directed into the cell. (a) If a pore opens, which way do sodium ions (Na") flow- into the cell or out of the cell? (b) What is the magnitude of the electric force on the sodium ion? The charge on the sodium ion is \(+e\)
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