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Chapter 21: Q6P (page 897)

Figure 21.61 shows disk shaped region of radius of 2 cm on which there is a uniform electric field of magnitude 300 V/m at an angle of 300 to the plane of the disk. Assume that points upward in +y direction. Calculate the electric flux on the disk, and include the correct units.

Short Answer

Expert verified

The electric flux for plane is 0.33V.m.

Step by step solution

01

Identification of given data

The radius of disk shaped region isr=2cm

The magnitude of uniform electric field is E=300V/m.

The angle of electric field with plane of disk isθ=30° .

02

Conceptual Explanation

The electric flux is calculated by the product of magnitude of electric field with the horizontal component of the surface area of disk.

03

Determination of electric flux of the disk

The surface area of the disk is given as:

A=πr2

Substitute all the values in the above equation.

A=π2cm1m100cm2A=1.26×10-3m2

The electric flux for plane is given as:

ϕ=EAcosθ

Substitute all the values in the above equation.

ϕ=300V/m1.26×10-3m2cos30°ϕ=0.33V·m

Therefore, the electric flux for plane is 0.33V.m.

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Most popular questions from this chapter

Figure 21.70 shows a close-up of the central region of a capacitor made of two large metal plates of area , very close together and charged equally and oppositely. There are +Q and -Q on the inner surfaces of the plates and small amounts of charge +q and -q on the outer surfaces.

(a). Knowing Q, determine E: Consider a Gaussian surface in the shape of a shoe box, with one end of area Abox in the interior of the left plate and the other end in the air gap (surface 1 in the diagram). Using only the fact that the electric field is expected to be horizontal everywhere in this region, use Gauss’s law to determine the magnitude of the field in the air gap. Check that your result agrees with our earlier calculations in the Chapter 15. (Be sure to consider the flux on all faces of your Gaussian box.)

The electric field has been measured to be horizontal and to the right everywhere on the closed box as shown in Figure 21.66. All over the left side of box E1=100V/m and all over the right (slanting) side of box E2=300V/m.On the top the average field is E3=150V/m , on the front and back the average field is E4=175V/mand on the bottom the average field is E5=220V/m.How much charge is inside the box? Explain briefly.

Question: A negative point charge –Q is at the center of a hollow insulting spherical shell, which has an inner radius R1 and an outer radius R2. There is a total charge of +3Q spread uniformly throughout volume of insulating shell, not just on its surface. Determine the electric field for (a) r<R1 (b) R1<r<R2 (c) R2<r.

The electric field is horizontal and has the values indicated on the surface of cylinder as shown in Figure 21.65. What can you deduce from this pattern of electric field? Include a numerical result.

A lead nucleus is spherical with a radius of about 7×10-15m . The nucleus contains 82protons (and typically126 neutrons). Because of their motions the protons can be considered on average to be uniformly distributed throughout the nucleus. Base on the net flux at the surface of the nucleus, calculate the divergence of the electric field as electric flux per unit volume. Repeat the calculation at a radius of3.5×10-15m . (You can use Gauss’s law to determine the magnitude of the electric field at this radius.) Also calculate the quantityρ/ε0 inside the nucleus.

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