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Chapter 25: Q. 9 (page 708)

FIGUREQ25.9shows two points inside a capacitor. LetV=0V at the negative plate.
aWhat is the ratioV2V1 of the electric potentials? Explain.
bWhat is the ratioE2E1of the electric field strengths?

Short Answer

Expert verified

Part a

aThe ratio of eletric potentials is V2/V1=Es2/Es1=3.

Partb

bThe ratio of eletric field strength isE2/E1=1.

Step by step solution

01

Step: 1 Capacitor at negative plate:

Electrical capacitance is measured in Farads and is represented as a ratio of charge to potential difference. The electric charge on each conductor is directly proportional to the potential between them, and this is how capacitance is calculated.

02

Step: 2 Ratio of electric potential: (part a)

Electrical capacitance is measured in Farads and is defined as the ratio of charge to potential difference. The ratio of the electromotive force on each circuit to the potential difference between them is known as capacitance.V2/V1=Es2/Es1=3.

03

Step: 3 Ratio of electric field strength: (part b)

The parallel plate capacitor is constant inside field,

E=ηϵ0,E2/E1=1.

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