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Chapter 26: Q. 12 (page 738)

Figure EX26.12 is a graph of Vversusx. Draw the corresponding graph of Exversusx.

Short Answer

Expert verified

We've drawn the required graph with slopes5000V/mand-5000V/m

Step by step solution

01

Given information and theory used  

Given :

Theory used :

The electric field is expressed as a function of potential as :

E(x)=-dV(x)dx

02

Drawing the corresponding graph of Ex versus x. 

The potential graph in our situation consists of two lines. As a result, the derivative will be a constant, namely the line's slope. The slopes can be seen to be steep.

5000V/mand-5000V/mare the slopes when drawing the graph. As a result, our graph would look like this:

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Most popular questions from this chapter

An infinitely long cylinder of radius Rhas linear charge density I. The potential on the surface of the cylinder isV0, and the electric field outside the cylinder is localid="1648568519666" Er=λ/2π∈0r. Find the potential relative to the surface at a point that is distance r from the axis, assuming localid="1648568540457" r>R.

Metal sphere 1 has a positive charge of 6.0 nC. Metal sphere 2, which is twice the diameter of sphere 1, is initially uncharged. The spheres are then connected together by a long, thin metal wire. What are the final charges on each sphere?

A capacitor being charged has a current carrying charge to and away from the plates. In the next chapter we will define current to be dQ/dt the rate of charge flow. What is the current to a 10μF capacitor whose voltage is increasing at the rate of 2.0V/s?

What are the magnitude and direction of the electric field at the dot in Figure EX26.7?

Estimate the electric fields E→1and E→2 at points 1 and 2 in Figure Q26.4. Don’t forget thatE→ is a vector.
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