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A rod with uniformly distributed charge 2×10-8C is 50cm long. We need to calculate E at a distance of 1cm from the midpoint of the rod. Which equation for the electric field of a rod should we use? (1) Exact, (2) Approximate, (3) Either exact or approximate, (4) Neither—we have to do it numerically, (5) Neither—we need to integrate.

Short Answer

Expert verified

The approximate equation can be used.

Step by step solution

01

Identification of given data

The charge on the rod is q=2×10-8C

The length of the rod is L=50cm

The distance of the measurement point from the rod is r=1cm

02

Approximation of electric field due to a charged rod

The approximated form of the electric field due to a charged rod on its mid plane is used when the distance of the point where the field is to be obtained from the rod is negligible compared to the length of the rod.

03

Determination of the method to obtain the field

Since the length of the rod L=50cm is much greater than the distance of the measurement point r=1cm , that is

Lr

the approximate form of the electric field formula can be safely used.

Hence, the approximate equation can be used.

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Most popular questions from this chapter

Two rings of radius 2 cm are 20 cm apart and concentric with a common horizontal x axis. What is the magnitude of the electric field midway between the rings if both rings carry a charge of +35 nC?

Question: Breakdown field strength for air is roughly . If the electric field is greater than this value, the air becomes a conductor. (a) There is a limit to the amount of charge that you can put on a metal sphere in air. If you slightly exceed this limit, why would breakdown occur, and why would the breakdown occur very near the surface of the sphere, rather than somewhere else? (b) How much excess charge can you put on a metal sphere of radius without causing breakdown in the neighboring air, which would discharge the sphere? (c) How much excess charge can you put on a metal sphere of onlyradius? These results hint at the reason why a highly charged piece of metal tends to spark at places where the radius of curvature is small, or at places where there are sharp points.

A solid metal ball of radius 1.5 cm bearing a charge of −17 nC is located near a solid plastic ball of radius 2 cm bearing a uniformly distributed charge of +7 nC (Figure 15.62) on its outer surface. The distance between the centers of the balls is 9 cm. (a) Show the approximate charge distribution in and on each ball. (b) What is the electric field at the center of the metal ball due only to the charges on the plastic ball? (c) What is the net electric field at the center of the metal ball? (d) What is the electric field at the center of the metal ball due only to the charges on the surface of the metal ball?

For a disk of radius R=20cm and Q=6×10-6C, calculate the electric field 2 mm from the center of the disk using all three equations:

role="math" localid="1656928965291" E=(Q/A)2ε0[1-z(R2+z)1/2]

EQ/A2e0[1-zR],andEQ/A2e0

How good are the approximate equations at this distance? For the same disk, calculate E at a distance of 5 cm (50 mm) using all three equations. How good are the approximate equations at this distance?

A large, thin plastic disk with radiusR = 1.5 m carries a uniformly distributed charge of −Q = −3 × 10−5 C as shown in Figure 15.59. A circular piece of aluminum foil is placed d = 3 mm from the disk, parallel to the disk. The foil has a radius of r = 2 cm and a thickness t = 1 mm.


(a) Show the charge distribution on the close-up of the foil. (b) Calculate the magnitude and direction of the electric field at location × at the center of the foil, inside the foil. (c) Calculate the magnitude q of the charge on the left circular face of the foil.

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