What is the electric flux through the surface shown in FIGURE EX24.9?

Short Answer

Expert verified

The electric flux through the surface is-2.3N·m²/C

Step by step solution

01

Given information and Theory used

Given figure :

Theory used :

The quantity of electric field passing through a closed surface is known as the Electric flux. Gauss's law indicates that the electric field across a surface is proportional to the angle at which it passes, hence we can determine charge inside the surface using the equation below.

Φe=E·A·cosθ

Where θis the angle formed between the electric field and the normal.

02

Calculating the electric flux

The electric field is uniform here, and the sheet is tilted to the electric field by an angle of 30°, hence we use equation (1) to calculate flux.

As indicated in the diagram, θmay be calculated as :

θ=90°+30°=120°

On the flat sheet, where θdoes not change and Ais the area of the flat sheet, which we can get by :

A=(0.15m×0.15m)=2.25x10-2m²

Plugging in our values for E,Aandθinto equation (1) to get the electric flux :

Φe=EAcosθ=(180N/C)(2.25x10-2m²)(cos120°)=-2.3N·m²/C

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Most popular questions from this chapter

The three parallel planes of charge shown in FIGURE P24.45have surface charge densities -12,h,handlocalid="1649410735638" -12,h- . Find the electric fields localid="1649410752965" Eu1to localid="1649410757308" Eu4in regions localid="1649410763257" 1to localid="1649410765846" 4.

Suppose you have the uniformly charged cube in FIGURE Q24.1. Can you use symmetry alone to deduce the shape of the cube’s electric field? If so, sketch and describe the field shape. If not, why not?

Newton’s law of gravity and Coulomb’s law are both inversesquare laws. Consequently, there should be a “Gauss’s law for gravity.” a. The electric field was defined as E u = F u on q /q, and we used this to find the electric field of a point charge. Using analogous reasoning, what is the gravitational field g u of a point mass?

Write your answer using the unit vector nr, but be careful with signs; the gravitational force between two “like masses” is attractive, not repulsive. b. What is Gauss’s law for gravity, the gravitational equivalent of Equation 24.18? Use ΦG for the gravitational flux, g u for the gravitational field, and Min for the enclosed mass. c. A spherical planet is discovered with mass M, radius R, and a mass density that varies with radius as r = r011 - r/2R2, where r0 is the density at the center. Determine r0 in terms of M and R. Hint: Divide the planet into infinitesimal shells of thickness dr, then sum (i.e., integrate) their masses. d. Find an expression for the gravitational field strength inside the planet at distance r 6 R.

A sphere of radius Rhas total charge Q. The volume charge density C/m3within the sphere is

ρ=ρ01-rR

This charge density decreases linearly from \(\rho_{0}\) at the center to zero at the edge of the sphere.

a. Show that ρ0=3Q/πR3.

b. Show that the electric field inside the sphere points radially outward with magnitude

E=Qr4πϵ0R34-3rR

| A spherical ball of charge has radius R and total charge Q. The electric field strength inside the ball 1r … R2 is E1r2 = r4 Emax /R4 . a. What is Emax in terms of Q and R? b. Find an expression for the volume charge density r1r2 inside the ball as a function ofr.c. Verify that your charge density gives the total charge Q when integrated over the volume of the ball

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