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Electrostatic Potential and Capacitance

Electric flux

A cylinder of height ' H ' and radius of cross-section ' R ' i

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Solving time: 3 mins

A cylinder of height ' $H$ ' and radius of cross-section ' $R$ ' is kept in uniform electric field $E$ as shown in figure. The magnitude of flux that enters from left half of cylinder is

E π R^{2}

ER H

2 E RH

Zero

Views: 8,563 students

Updated on: Jul 17, 2023

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Olivia Discussed

A cylinder of height '

H

' and radius of cross-section '

R

' is kept in uniform electric field

E

as shown in figure. The magnitude of flux that enters from left half of cylinder is

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

As in the given diagram, electric field is uniform.
So, electric flux

=

Projected area

\times

Electric field
Electric flux

= (2 R H) E

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Question Text	A cylinder of height ' $H$ ' and radius of cross-section ' $R$ ' is kept in uniform electric field $E$ as shown in figure. The magnitude of flux that enters from left half of cylinder is
Updated On	Jul 17, 2023
Topic	Electrostatic Potential and Capacitance
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 5
Upvotes	423
Avg. Video Duration	4 min

A cylinder of height ' $H$ ' and radius of cross-section ' $R$ ' is kept in uniform electric field $E$ as shown in figure. The magnitude of flux that enters from left half of cylinder is

Text solutionVerified

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A cylinder of height ' H ' and radius of cross-section ' R ' is kept in uniform electric field E as shown in figure. The magnitude of flux that enters from left half of cylinder is

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A cylinder of height ' $H$ ' and radius of cross-section ' $R$ ' is kept in uniform electric field $E$ as shown in figure. The magnitude of flux that enters from left half of cylinder is