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The electric field on a closed surface is due to all the charges in the universe, including the charges outside the closed surface. Explain why the total flux nevertheless proportional only to the charges that are inside the surface, with no apparent influence of the charges outside.

Short Answer

Expert verified

The electric flux is proportional to inside charge of enclosed surface but not the outside charges because electric field lines are always normal to the enclosed surface for inside charges.

Step by step solution

01

Conceptual Explanation

The electric flux is the effect of the electric field lines passing through the certain cross section of an area. The effect varies with the orientation of cross section with electric field lines

02

Determination of Electric flux

The electric flux due to any charge or system of charges is given as:

ϕ=E·Aϕ=EAcosθ

Here, E is the intensity of electric field lines due the charges, A is the area of cross section and θ is the direction between electric field lines and area of cross section from where electric field lines passes.

The value of electric field lines varies with the direction between electric field lines and area of cross section. The flux is maximum for normal electric field lines to the surface and zero for the tangential electric field lines to the surface.

03

Reason for electric flux due to enclosed charge proportional to inside charge not for outside charges

The electric filed lines for inside charges of enclosed surface are normal to the surface so the electric flux is proportional to inside charges. The electric field lines for outside charges of enclosed surface are tangential to the surface so the electric flux has different values for different directions.

Therefore, the electric flux is proportional to inside charge of enclosed surface but not the outside charges because electric field lines are always normal to the enclosed surface for inside charges.

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Most popular questions from this chapter

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.

Based on its definition as electric flux per unit volume, what are the units of the divergence of electric field.

The center of a thin paper cube in outer space is located at the origin. Each edge is 10 cm long. The only other objects in the neighborhood are some small charged particles whose charges and position at this instant are following: +9 nC at (4,2,-3) cm, -6 nC at (1,-3,2) cm, +7 nC at (15,0,4) cm, +4 nC at (-1,3,2) cm and -3 nC at (-2,12,-3) cm. The electric flux on the five of the six faces of the cube totals 564 V m. What is the flux on the other face?

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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