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Chapter 26: Q. 7 (page 737)

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

Short Answer

Expert verified

The magnitude is 20000V/m and direction of the electric field isvertically downwards.

Step by step solution

01

Given information and formula used  

Given figure :

Theory used :

The electric field is directed towards the decreasing potential and has a magnitude equal to the rate of change of the potential.

The electric field will be equal to the potential difference divided by the distance between these potentials. That is to say,

E=鈭�Vd

02

Determining the magnitude and direction of the electric field at the dot in the figure

The potential varies by200Vover a distance of1cm=0.01min our case, hence the field's magnitude is :

E=200V0.01m=20000V/m

The field points downwards as the potential falls in the downward direction.

So, the Electric field points vertically downwards.

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Most popular questions from this chapter

Figure Q26.10 shows a 3Vbattery with metal wires attached to each end. What are the potential differences 鈭�V12=V2-V1,鈭�V23=V3-V2,鈭�V34=V4-V3,and鈭�V41=V1-V4?

How much work does the charge escalator do to move 1.0渭C of charge from the negative terminal to the positive terminal of a 1.5V battery?

Figure Q26.1 shows the x-component of E鈫�as a function of x. Draw a graph of Vversusxin this same region of space. Let V=0Vatx=0mand include an appropriate vertical scale.

a. Which point in Figure EX26.5, 'A' or 'B', has a larger electric potential?

b. What is the potential difference between 'A' and 'B'?

An isolated 5.0渭F parallel-plate capacitor has 4.0mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.0渭F. How much work is done by the external force?
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