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Equation 23-11 (E=σ/ε0) gives the electric field at points near a charged conducting surface. Apply this equation to a conducting sphere of radius rand charge q, and show that the electric field outside the sphere is the same as the field of a charged particle located at the center of the sphere.

Short Answer

Expert verified

The electric field outside the sphere is the same as the field of a charged particle located at the center of the sphere.

Step by step solution

01

Listing the given quantities

The electric field is given by,E=δ/ε0

02

Understanding the concept of electric field

We interpret the question as referring to the field just outside the sphere (that is, at locations roughly equal to the radius r of the sphere). Since the area of a sphere A=4πr2is and the surface charge densityδ=q/Ais (where we assume q is positive for brevity),

Formula:

E=δε0

03

Deriving the expression for electric field

The electric field at the points near the charged conducting surface is given by,

E=δε0

To apply this equation to the conducting sphere to find the electric field, let’s first find the charge density due to conducting sphere of radius and charge ,

δ=qA

The area of the sphere can be written in terms of radius as,

A=4πr2

Substitute the value in the above equation,

δ=q4πr2

Now, use the value of charge density to calculate the electric field as,

E=1ε0q4πr2=q4πε0r2

The above expression is the same as the expression for the electric filed field of a point charge.

Therefore, the electric field outside the sphere is the same as the field of a charged particle located at the center of the sphere.

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Most popular questions from this chapter

A particle of charge+q is placed at one corner of a Gaussian cube. What multiple ofq/ε0gives the flux through (a) each cube face forming that corner and (b) each of the other cubes faces?

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At each point on the surface of the cube shown in Fig. 23-31, the electric field is parallel to the z-axis. The length of each edge of the cube is3.0m. On the top face of the cube the field is E=-34k^N/Cand on the bottom face it isE=+20k^N/C Determine the net charge contained within the cube.

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A Gaussian surface in the form of a hemisphere of radiusR=5.68cmlies in a uniform electric field of magnitudeE=2.50N/C. The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. What is the flux through

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