A point charge of 3×109 Cis located at the origin.

(a) What is the magnitude of the electric field at location 0.2,0,0 m?

(b) Next, a short, straight, thin copper wire 3 mmlong is placed along the x axis with its center at location 0.1,0,0 m. What is the approximate change in the magnitude of the electric field at location 0.2,0,0 m?

(c) Does the magnitude of the electric field at location 0.2,0,0 m increase or decrease as a result of placing the copper wire between this location and the point charge?

(d) Does the copper metal block the electric field contributed by the point charge?

Short Answer

Expert verified
  1. The magnitude of the electric field at location 0.2,0,0 m is 675 N/C.
  2. The approximate change in the magnitude of the electric field at location 0.2,0,0 m is 9.1×103 N/C.
  3. The magnitude of the electric field at location 0.2,0,0 m increases.
  4. The point charge does not contribute to the electric field of the copper metal block.

Step by step solution

01

Identification of the given data

The given data can be listed below as:

  • The charge of the point charge is Q=3×109 C.
  • The location of the electric field is, r=0.2,0,0 m.
  • The length of the copper wire is, s=3 mm×10-3 m1 mm=3×103 m.
  • The location of the copper wire is, d=0.1,0,0 m.
02

Significance of the magnitude of the electric field due to dipole and the point charge

The magnitude of the electric field due to a point charge is directly proportional with the charge and inversely proportional to the square of the distance of the charge to the center of the electric field.

The magnitude of the electric field due to a dipole is directly proportional with the charge of the dipole and the distance of separation and inversely proportional to the cube of the distance of the dipole to the center of the electric field.

03

(a) Determination of the magnitude of the electric field

The equation of the magnitude of the location of the electric field can be expressed as:

r=rx2+ry2+rz2

Here, r is the magnitude of the location of the electric field, rxis the location of the electric field at the xaxis, ryis the location of the electric field at the y axis and rz is the location of the electric field at the role="math" localid="1661327540844" zaxis.

Substitute the values in the above equation.

r=(0.2)2+(0)2+(0)2 m=0.2 m

The equation of the magnitude of the electric field due to a point charge is expressed as:

E=kQr2

Here, E is the magnitude of the electric field due to a point charge, k is the electric field constant, Q is the point charge and r is the magnitude of the location of the electric field.

Substitute the values in the above equation.

E=(9×109 Nm2/C2)3×10-9 C(0.2 m)2=18 Nm2/C(0.04 m2)=675 N/C

Thus, the magnitude of the electric field at location 0.2,0,0 m is 675 N/C.

04

(b) Determination of the approximate change in the magnitude of the electric field

As the wire behaves like a conductor, at equilibrium, the net electric field is zero. The equation of the net electric field can be expressed as:

Enet=Eext+Epol=0

Here, Enet is the magnitude of the net electric field, Eextis the magnitude of the external electric field and Epol is the magnitude of the polar electric field.

The above equation can also be expressed as:

0=Eext+Enegative plate+Epositive plate …(i)

Here, Eext is the magnitude of the external electric field and Enegative plateis the magnitude of the electric field of the negative plate and Epositive plate is the magnitude of the electric field of the positive plate.

The equation of the magnitude of the external electric field is expressed as:

Eext=kQr2 …(ii)

Here, Eext is the magnitude of the external electric field due to a point charge, kis the electric field constant, Qis the point charge and ris the magnitude of the location of the electric field.

The equation of the magnitude of the negative plate electric field is expressed as:

Enegative plate=kQ1(s/2)2 … (iii)

Here, Enegative plate is the magnitude of the electric field of the negative plate, k is the electric field constant, Q1is the charge of the induced dipole and sis the length of the copper wire.

The equation of the magnitude of the positive plate electric field is expressed as:

Epositive plate=kQ1(s/2)2 …(iv)

Here, Epositive plate is the magnitude of the electric field of the positive plate, k is the electric field constant, Q1 is the charge of the induced dipole and sis the length of the copper wire.

Substitute the values of the equation (ii), (iii) and (iv) in the equation (i).

0=Qr2kQ1(s/2)2kQ1(s/2)2Qr2=2Q1(s/2)2Q1=Q8sr2

The equation of the change in the magnitude of the electric field is expressed as:

E1=kQ1sr3=ksr3Q8sr2

Here, E1 is the change in the magnitude of the electric field, k is the electric field constant, ris the magnitude of the location of the electric field and s is the length of the copper wire.

Substitute the values in the above equation.

E1=(9×109 Nm2/C2)(3×10-3 m)(0.2 m)33×109 C83×10-3 m0.1 m2=(27×106 Nm3/C2)(8×103 m3)3×109 C9×10-6 m20.01 m2=(3.3×109 N/C2)2.7×1012 C9.1×103 N/C

Thus, the approximate change in the magnitude of the electric field at location 0.2,0,0 m is 9.1×103 N/C.

05

(c) Determination of the increase or decrease of the magnitude of the electric field

The change in the magnitude due to the copper wire at location 0.2,0,0 m is 9.1×103 N/C. Hence, it has been identified that the magnitude of the electric field increases as it consists that value.

Thus, the magnitude of the electric field at location 0.2,0,0 m increases.

06

(d) Determination of the contribution of the point charge on the copper block

The point charge is mainly helpful for finding the magnitude of the net electric field due to a point charge. The copper metal block has its own electric field and the point charge does not contribute in the electric field.

Thus, the point charge does not contribute to the electric field of the copper metal block.

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Most popular questions from this chapter

A glass sphere carrying a uniformly distributed charge of +Qis surrounded by an initially neutral spherical plastic shell (Figure 15.67).

(a) Qualitatively, indicate the polarization of the plastic. (b) Qualitatively, indicate the polarization of the inner glass sphere. Explain briefly. (c) Is the electric field at location P outside the plastic shell larger, smaller, or the same as it would be if the plastic weren’t there? Explain briefly. (d) Now suppose that the glass sphere carrying a uniform charge of +Qis surrounded by an initially neutral metal shell (Figure 15.68). Qualitatively, indicate the polarization of the metal.

e) Now be quantitative about the polarization of the metal sphere and prove your assertions. (f) Is the electric field at location P outside the metal shell larger, smaller, or the same as it would be if the metal shell weren’t there? Explain briefly.

A very thin spherical plastic shell of radius15 cm carries a uniformly distributed negative charge of 8 nC(8×109 C)on its outer surface (so it makes an electric field as though all the charge were concentrated at the center of the sphere). An uncharged solid metal block is placed nearby. The block is10cm thick, and it is10cm away from the surface of the sphere. See Figure 14.97. (a) Sketch the approximate charge distribution of the neutral solid metal block.

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A student said, “When you touch a charged piece of metal, the metal is no longer charged: all the charge on the metal is neutralized.” As a practical matter, this is nearly correct, but it Isn’t exactly right. What’s wrong with saying that all the charge on the metal is neutralized?

A small glass ball is rubbed all over with a small silk cloth and acquires a charge of +5nC. The silk cloth and the glass ball are placed 30 cm apart.

(a) On a diagram like that shown in Figure 14.71, draw the electric field vectors qualitatively at the locations marked ×. Pay careful attention to directions and to relative magnitudes. Use dashed lines to explain your reasoning graphically, and draw the final electric field vectors with solid lines.

(b) Next, a neutral block of copper is placed between the silk and the glass.

On a diagram like that shown in Figure 14.72, carefully show the approximate charge distribution for the copper block and the electric field vectors inside the copper at the locations marked ×.

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(d) Even if you have to state your result as an inequality, make as quantitative a statement as you can about the electric field at the location of the glass ball and the net force on the ball when the plastic block is in place compared to when there is no block. Explain briefly.

A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1×1010excess electrons. An initially uncharged identical metal ball hanging from an insulating thread is brought in contact with the first ball, then moved away, and they hang so that the distance between their centers is 20cm.

(a) Calculate the electric force one ball exerts on the other, and state whether it is attractive or repulsive. If you have to make any simplifying assumptions, state them explicitly and justify them.

(b) Now the balls are moved so that as they hang, the distance between their centers is only 5cm. Naively one would expect the force that one ball exerts on the other to increase by a factor of 42=16, but in real life the increase is a bit less than a factor of role="math" localid="1661330186132" 16. Explain why, including a diagram. (Nothing but the distance between centers is changed—the charge on each ball is unchanged, and no other objects are around.)

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