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Chapter 26: Q. 6 (page 737)

Two flat, parallel electrodes 2.5cmapart are kept at potentials of20Vand35V. Estimate the electric field strength between them.

Short Answer

Expert verified

The electric field strength is600V/m

Step by step solution

01

Given information and formula used  

Given :

Distance between the electrodes : 2.5cm

Are kept at potentials of : 20Vand35V

Theory used :

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

Calculating the electric field strength between the two parallel electrodes 

Since we have : E=鈭�Vd

So, in our numerical example,

E=35-200.025=600V/m

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Most popular questions from this chapter

a. Use the methods of Chapter 25 to find the potential at distance xon the axis of the charged rod shown in FIGURE P26.43.

b. Use the result of part a to find the electric field at distance xon the axis of a rod

An electron is released from rest atx=2m in the potential shown in Figure Q26.6. Does it move? If so, to the left or to the right? Explain.

A switch that connects a battery to a 10 mF capacitor is closed. Several seconds later you find that the capacitor plates are charged to 卤30 mC. What is the emf of the battery?

The parallel-plate capacitor in Figure Q26.11 is connected to a battery having potential difference 鈭�Vbat. Without breaking any of the connections, insulating handles are used to increase the plate separation to 2d.

a. Does the potential difference 鈭�VCchange as the separation increases? If so, by what factor? If not, why not?

b. Does the capacitance change? If so, by what factor? If not, why not?

c. Does the capacitor charge Qchange? If so, by what factor? If not, why not?

Determine the magnitude and direction of the electric field at points 1 and 2 in Figure EX26.10.
See all solutions

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