Motion of charged particle in electric field

Electricity and Magnetism Level 2

In the above diagram, a charged particle is moving in an electric field, where the dotted red lines are equipotential lines. The charged particle is initially entered with a speed of $v$ from the left of point A . Then it passes points A and B , finally arriving at point C . It stops for a moment there at point C , and then starts to move back to point A . Which of the following statements is correct?

a) The charged particle is negatively charged.
b) The magnitude of the kinetic energy when the charged particle passes point B is half the kinetic energy when the charged particle passes point A .
c) The force acting on the charged particle at point C is not zero.

b) only b) and c) a) and b) a) only

4 solutions

Kazi Mifta
Apr 6, 2014

The correct answer is b and c. The answer lies in the question. As the particle went back after reaching point a it does meant that some kind of opposite force is acting. So the particle is positively charged. Again it is attracted at point A by the negative charge so the kinetic energy becomes half. And at point C a repulsion force acts.

the motion of the charged particle resembles that of a pendulum where the kinetic energy is max at the mean position and a restoring force acts at the extreme position.

shuvam jaiswal - 7 years ago

Sophie Crane
Apr 21, 2014

c) has to be true because of Newton's first law: if the force were zero, the particle would not stop and then move backwards again.

Ashik Shajahan
Mar 28, 2014

Frequency depends on distance..it gives d answer

the electrostatic potential at c is zero......in the figure....how is it possible????

Sayam Chakravarty - 7 years, 2 months ago

Did you mean at B? Why wouldn't it be possible? Also, it isn't like every theoretical problem in physics is taken from real scenarios. Some are purely theoretical to understand certain concepts. There are so many problems where you have to ignore friction or air resistance, for example.

Akshay Jamwal - 7 years, 2 months ago

e.p. at c is 6 v

Sanjay Bissu - 7 years, 2 months ago

how force at c is zero?

Durga Sadasivuni - 7 years, 2 months ago

force at c is not zero give in problem

Sanjay Bissu - 7 years, 2 months ago

Ravindhar Rockss - 7 years, 2 months ago

I think the answer is only C. Initial kinetic energy of the particle is KE. Let us say it is 10 units . As the particle moves from A to B it looses energy. This loss in energy is due to the work done on it as it moves from lower to higher potential. Let this work be 6 units ( assuming a unit charge ). Then the kinetic energy at B is 4 units ( which is not half the initial KE )

vijaykiran vemuri - 7 years, 2 months ago

at pnt a there is attraction btwn moving charge (with v) & steady charge which is against the motion and at c there is repulson which is acting in opposite to motion & in given problem charge stops at c so seeing this there is lost of half ke at pnt b due to that moving charge stops at c

Sanjay Bissu - 7 years, 2 months ago

It is understood that there is loss in energy but, how can we say EXACTLY HALF !

vijaykiran vemuri - 7 years, 2 months ago

@Vijaykiran Vemuri – From the figure, we can consider or ASSUME that the equipotentials are due to resultant of two equal charges as the equipotentials seem to be symmetric about B and potential at B=0 V... so clearly B is midway between the two charges.... suppose the charge on the particle is 'q'....since conservation of energy holds true here so...total energy(E) at "A'=K.E(A)+(-6)q=E total energy at 'B'=K.E(B)+0q=K.E(B)=E total energy at 'C'=0+6q=E from this we have K.E(A)=12q... K.E(B)=6q which is half of K.E at A

Chingangbam Dinesh - 7 years, 2 months ago

Can anybody please explain how option (b) is correct. I could only decipher about option (c).

Archit Gupta - 7 years, 2 months ago

In reply to Vijaykiran Vemuri: at pnt a there is attraction btwn moving charge (with v) & steady charge which is against the motion and at c there is repulson which is acting in opposite to motion & in given problem charge stops at c so seeing this there is lost of half ke at pnt b due to that moving charge stops at c

Sanjay Bissu - 7 years, 2 months ago

hte electronic potential at is c zero ...... the figure howw to possible ?????

john sanchez - 7 years, 2 months ago

the option given is force at c is non zero

Krishna Gudi - 7 years, 2 months ago

the charge of particle may be positive

Chintan Chawda - 7 years, 2 months ago

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Natalie Wignall - 7 years, 2 months ago

Subash Malik
Apr 5, 2014

answer is : b) and c)

kindly justify option c

zalmey khan - 7 years, 2 months ago

if force at c is zero than moving charge wouldnt stop at c due to its KE soforce at c is not zero

Sanjay Bissu - 7 years, 2 months ago

got it 'thanks sanjay

zalmey khan - 7 years, 2 months ago

C) is confirmed as it ws given tht d particle moved towards a .....it cn either be possible that point c gives d particle a repulsive force or it might be the attractive force by point a on d particle

Samarth Rastogi - 7 years, 2 months ago

Motion of charged particle in electric field

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