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FIGURE shows three Gaussian surfaces and the electric flux through each. What are the three charges q1,q2andq3?

Short Answer

Expert verified

The three charges, of gaussian surface and electric flux,q1=2q, q2=q, q3=-3q

Step by step solution

01

Derive the equations

The amount of electric field that travels through a covered surface is referred to as the electric flux. The electric flux through with a surface is proportional to the charge inside the surface, according to Gauss's law, which is given by the equation.

Φe=E×dA=Qinϵo

the electrical flux depends on the charge inside the closed surface. any flux due to charges outside the closed surface is zero

Pertaining to the enclosed surface,

ΦA=-qϵo, As a result, the surface's net charge is

q1+q3=-q (1)

pertaining to the enclosed surface,

ΦB=3qϵo, As a result, the surface's net charge is

q1+q2=3q (2)

pertaining to the enclosed surface,

ΦC=-2qϵo, As a result, the surface's net charge is

q3+q2=-2q (3)

02

Rearrange the equation to get the value

Rewrite the equation (2) in a different way

q1=3q-q2

Use this in equation (1)

q1+q3=-q

3q-q2+q3=-q

q3-q2=-4q (4)

Add equation (3)and (4)to get q3

q3+q2=-2q

+

q3-q2=-4q

2q3=-6q

q3=-3q

03

Derive the other equations

Put values of q3in equation (1)

q1=-q-q3=-q-(-3q)=2q

Use values of q1in equation (2)

q2=3q-q1=3q-2q=q

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Most popular questions from this chapter

FIGURE P24.48shows two very large slabs of metal that are parallel and distance lapart. The top and bottom of each slab has surface area A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal 1has total charge Q1=Qand metal 2has total charge Q2=2Q. Assume Qis positive. In terms of Qand A, determine

a. The electric field strengths E1toE5in regions 1to 5.

b. The surface charge densities ηuto ηdon the four surfaces a to d.

The electric field is constant over each face of the cube shown in FIGURE EX24.5. Does the box contain positive charge, negative charge, or no charge? Explain.

55.3million excess electrons are inside a closed surface. What is the net electric flux through the surface?

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/ϵ0,with Qin/ϵ1, where ϵis the permittivity of the material. (Technically, ϵ0is called the vacuum permittivity.) Suppose a long, straight wire with linear charge density 250nC/mis covered with insulation whose permittivity is 2.5ϵ0. What is the electric field strength at a point inside the insulation that is 1.5mmfrom the axis of the wire?

A small, metal sphere hangs by an insulating thread within the larger, hollow conducting sphere of FIGURE Q24.10. A conducting wire extends from the small sphere through, but not touching, a small hole in the hollow sphere. A charged rod is used to transfer positive charge to the protruding wire. After the charged rod has touched the wire and been removed, are the following surfaces positive, negative, or not charged? Explain. a. The small sphere. b. The inner surface of the hollow sphere. c. The outer surface of the hollow sphere.

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