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What is wrong with Figure 15.35 and this associated incorrect student explanation? “The electric field at location inside the uniformly charged sphere points in the direction shown, because the charges closest to this location have the largest effect.” (Spheres provide the most common exception to the normally useful rule that the nearest charges usually make the largest contribution to the electric field.)

Short Answer

Expert verified

The wrong fact about the given figure is the direction of the electric field at the location that is directed outward, but the electric field is zero.

Step by step solution

01

Significance of the electric field

The electric field is referred to as a region which allows a charged particle to exert a particular force on another charged particle.

The equation of the electric field gives the answer that is wrong with the figure.

02

Identification of the problem with the figure

It can be observed from the given figure that the negative charge outside the sphere is uniformly distributed and the other negative charges are further away, due to the excessive amount of the negative charges. The electric field at the location and the negative charges outside the spheres cancels each other. So, the electric field becomes zero. Hence, the direction of the electric field cannot be outward as it has zero value.

Thus, the wrong fact about the given figure is the direction of the electric field at the location that is directed outward, but the electric field is zero.

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Most popular questions from this chapter

If the magnitude of the electric field in air exceeds roughly3×10-6N/C, the air breaks down and a spark forms. For a two-disk capacitor of radius50cmwith a gap of role="math" localid="1656068507772" 1mm, what is the maximum charge (plus and minus) that can be placed on the disks without a spark forming (which would permit charge to flow from one disk to the other)? Under these conditions, what is the strength of the fringe field just outside the center of the capacitor?

In a cathode-ray tube, an electron travels in a vacuum and enters a region between two deflection plates where there is an upward electric field of magnitude1×105N/C(Figure 15.60).


(a) Sketch the trajectory of the electron, continuing on well past the deflection plates (the electron is going fast enough that it does not strike the plates). (b) Calculate the acceleration of the electron while it is between the deflection plates. (c) The deflection plates measure 12 cm by 3 cm, and the gap between them is 2.5 mm. The plates are charged equally and oppositely. What are the magnitude and sign of the charge on the upper plate?

The rod in Figure 15.49 carries a uniformly distributed positive charge. Which arrow (a–h) best represents the direction of the electric field at the observation location marked with a red X?

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.

Suppose that the radius of a disk is 21 cm, and the total charge distributed uniformly all over the disk is 5×10-6C. (a) Use the exact result to calculate the electric field 1 mm from the center of the disk. (b) Use the exact result to calculate the electric field 3 mm from the center of the disk. (c) Does the field decrease significantly?

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