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Suppose an electric field E(x.y,z)has the form

Ex=ax,Ey=0,Ez=0

Where ais a constant. What is the charge density? How do you account for the fact that the field points in a particular direction, when the charge density is uniform?

Short Answer

Expert verified

Answer

The formula of charge density is =0[E].

Here, Eis linear function of x,yand z. From this it is clear that electric field points in particular direction.

Step by step solution

01

Define functions

Write the expression for electric filed from divergence theorem

E=e0 鈥︹ (1)

Here, Eis the electric filed, is the electric filed, 0is the permittivity for the free space.

02

Determine charge density

Write the formula for electric filed component along with x-axis.

E=zx^x 鈥︹ (2)

Write the expression for charge density.

=0[VE] 鈥︹ (3)

E=Exx+Eyx+E2x 鈥︹ (4)

The electric field component along y and z axis is zero. Therefore, Eis expressed as,

E=Exx 鈥︹ (5)

Substitute the value Exxfor Ein equation (3)

=0[Exx]

Differentiate Exand consider the value from equation (2),

Exx=x(ax)=a

Substitute afor Exxin equation (3)

=0a

From the above equation, it is clear that is constant everywhere.

Thus, the charge density is =Constant.

03

Determine charge density is uniform

Write the expression for charge density by equation (3)

=0E

From the above equation the charge density is directly proportional to the electric filed.

If charge density is uniform then,

E=Constant

Therefore, The values of Exx+Eyx+Ezx are also constant.

Since Eis linear function of x,yand z. From this it is clear that electric field points in particular direction.

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Most popular questions from this chapter

Imagine that new and extraordinarily precise measurements have revealed an error in Coulomb's law. The actual force of interaction between two point charges is found to be

F=14蟺蔚0q1q2r2(1+r)e(r)r^

where 位 is a new constant of nature (it has dimensions of length, obviously, and is a huge number鈥攕ay half the radius of the known universe鈥攕o that the correction is small, which is why no one ever noticed the discrepancy before). You are charged with the task of reformulating electrostatics to accommodate the new discovery. Assume the principle of superposition still holds.

a. What is the electric field of a charge distribution 蚁 (replacing Eq. 2.8)?

b. Does this electric field admit a scalar potential? Explain briefly how you reached your conclusion. (No formal proof necessary鈥攋ust a persuasive argument.)

c. Find the potential of a point charge q鈥攖he analog to Eq. 2.26. (If your answer to (b) was "no," better go back and change it!) Use 鈭 as your reference point.

d. For a point charge q at the origin, show that

SE.da+12V痴诲蟿=10q

where S is the surface, V the volume, of any sphere centered at q.

e. Show that this result generalizes:

SE.da+12V痴诲蟿=10Qenc

for any charge distribution. (This is the next best thing to Gauss's Law, in the new "electrostatics.鈥)

f. Draw the triangle diagram (like Fig. 2.35) for this world, putting in all the appropriate formulas. (Think of Poisson's equation as the formula for 蚁 in terms of V, and Gauss's law (differential form) as an equation for 蚁 in terms of E.)

g. Show that some of the charge on a conductor distributes itself (uniformly!) over the volume, with the remainder on the surface. [Hint: E is still zero, inside a conductor.]

A long coaxial cable (Fig. 2.26) carries a uniform volume charge density pon the inner cylinder (radius a ), and a uniform surface charge density on the outer cylindrical shell (radius b ). Thissurface charge is negative and is of just the right magnitude that the cable as a whole is electrically neutral. Find the electric field in each of the three regions: (i) inside the inner cylinder(s<a),(ii) between the cylinders(a<s<b)(iii) outside the cable(s>b)Plot lEI as a function of s.

A metal sphere of radiusRcarries a total chargeQ.What is the force

of repulsion between the "northern" hemisphere and the "southern" hemisphere?

Here is a fourth way of computing the energy of a uniformly charged

solid sphere: Assemble it like a snowball, layer by layer, each time bringing in aninfinitesimal charge dqfrom far away and smearing it uniformly over the surface,thereby increasing the radius. How much workdWdoes it take to build up the radius by an amountlocalid="1654664956615" dr? Integrate this to find the work necessary to create the entire sphere of radius Rand total charge q.

Find the energy stored in a uniformly charged solid sphere of radiusRand charge q.Do it three different ways:

(a)Use Eq. 2.43. You found the potential in Prob. 2.21.

(b)Use Eq. 2.45. Don't forget to integrate over all space.

(c)Use Eq. 2.44. Take a spherical volume of radiusa.What happens as a?

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