A proton is moved 15 cm on a path parallel to the field lines of a uniform electric field of 2 . 0 × 10 5 V m - 1 . What are the possible changes in potential? Consider both cases of moving the proton along and against the field.

Given, E = 2 . 0 × 10 5 N C - 1 , d = 15 × 10 - 2 Case I : When a proton is moved along the electric field. The proton is moved from A to B . Using the formula V = - E → · d → = - E d cos θ Here, θ is the angle between E and d , θ = 0 ° Thus, V A B = - E d × 1 Putting the values we get, V A B = - 3 . 0 × 10 4 volts Case II : When the proton is moved against the electric field. The proton is moved from B to A . Using the formula V = - E → · d → = - E d cos θ Here, θ is the angle between E and d , θ = 180 ° Thus, V A B = - E d × - 1 Putting the values we get, V A B = 3 . 0 × 10 4 volts .

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Beta Question Bank for Medical: Physics>Chapter 22 - Electrostatics>BEGINNER'S BOX - 5>Q 14

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Earn 100

What is the difference in potentials between two points, one at $10 cm$ and the other at $20 cm$ distance from a charge of $- 5.5 μ C$ ?

Important Questions on Electrostatics

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IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 22 - Electrostatics>BEGINNER'S BOX - 6>Q 1

Two points are there, one at $10 cm$ distance, and the other at $20 cm$ distance from a charge of $- 5.5 μC$ . Which point is at a higher potential?

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Beta Question Bank for Medical: Physics>Chapter 22 - Electrostatics>BEGINNER'S BOX - 6>Q 2

How far would a point be from a

+ 0.60 μC

, charge if it had a value of electric potential equal to

10 kV

?

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Beta Question Bank for Medical: Physics>Chapter 22 - Electrostatics>BEGINNER'S BOX - 6>Q 3

A point at some distance from a $+ 0.60 μC$ charge has a value of electric potential equal to $10 kV$ . If the point were moved to four times its initial distance from the charge then what potential change would occur? Is it an increase or a decrease?