Uniform distribution of charge and work needed

[fight_club_alum]

Homework Statement

A charge of +3.0 μC is distributed uniformly along the circumference of a circle with a
radius of 20 cm. How much external energy is required to bring a charge of 25μC from infinity to the centre of the circle?

a . 5.4 J

b. 3.4 J <- answer

c. 4.3 J

d. 2.7 J

e. 6.8 J=

Homework Equations

(delta) V = (kq/r1) + (kq2/r2) + (kq3/r3) + ...
work = q * delta v

The Attempt at a Solution

if we normally add the charges this way, then we should use the linear charge density to calculate how often the charge repeats
(lambda) = q/2pi*0.2
(Lamda) * (delta) v * (25*10^-6) = (not the answer)
Any help, please?

 

[kuruman]

What charges are you "normally" adding this way? First find the electric potential (relative to infinity) at the center of the ring. Then find the potential energy change of the 25 μC charge when it is moved from infinity to the center of the ring. How is that potential energy change related to the question of the problem?

 

[fight_club_alum]

If for instance we have a 1q =3uc charge and a 12 =5 uc charge the same distance from the centre, 0.2 m
I think we do so, (kq1/0.2) + (kq2/0.2)
Isn't that correct

 

[kuruman]

That is correct. What do you get when you add all these together?

 

[fight_club_alum]

That is correct. What do you get when you add all these together?

I get the potential at the centre (distance 0.2 m from both charges)

 

[kuruman]

Correct, but do you only have two charges? If so what are their values in terms of what is given to you?

 

[fight_club_alum]

That is correct. What do you get when you add all these together?

The thing is that I think that the two concepts are the same since we have many charges that are 3uc distributed uniformly along a circumference aren't we supposed to multiply the linear charge density time * (kq/r) (of only one since all charges are sharing the same distance but in different orientations) to get the total electric potentail

 

[fight_club_alum]

Correct, but do you only have two charges? If so what are their values in terms of what is given to you?

I have one charge but it is repeated along the circumference in a specific pattern (lambda)
Am i being correct?

 

[kuruman]

It looks like you have a blind spot and you don't see the obvious, so I will help you along. Suppose the Q = 3 μC were all concentrated on one spot spot at the circumference of radius $R$, say the 12 o' clock position. What is the electrostatic potential at the center? Answer: $V=\frac{kQ}{R}$. Suppose now that you split the charge into two equal pieces and placed one at 12 o' clock and one at 6 o' clock. What is the potential at the center now? Answer: $V=\frac{kQ}{2R}+\frac{kQ}{2R}=\frac{kQ}{R}$. Now put $\frac{1}{4}Q$ at four different positions on the circumference. What is is the potential at the center? What about if you took the initial charge $Q$, split it into infinitely many pieces and put it all on the circumference? Do you see what's going on?

 

[fight_club_alum]

It looks like you have a blind spot and you don't see the obvious, so I will help ypu along. Suppose the Q = 3 μC were all concentrated on one spot spot at the circumference of radius $R$, say the 12 o' clock position. What is the electrostatic potential at the center? Answer: $V=\frac{kQ}{R}$. Suppose now that you split the charge into two equal pieces and placed one at 12 o' clock and one at 6 o' clock. What is the potential at the center now? Answer: $V=\frac{kQ}{2R}+\frac{kQ}{2R}=\frac{kQ}{R}$. Now put $\frac{1}{4}Q$ at four different positions on the circumference. What is is the potential at the center? What about if you took the initial charge $Q$, split it into infinitely many pieces and put it all on the circumference? Do you see what's going on?

Great! I understand the point by now. I just thought that we will repeat the (kq/r) lambda times to get the total potential at the centre but your words make perfect sense
thank you very much, and to make sure I understand, if it was a quarter or half circle, it would be calculated the same way as a full circle?
Thank you so much again

 

[kuruman]

thank you very much, and to make sure I understand, if it was a quarter or half circle, it would be calculated the same way as a full circle?

Yes, and if you shrink the semi circle back to a point charge $Q$ = 3 μC at distance $R$ from the center, the potential will still be the same and you will be back where you started.

 

[fight_club_alum]

Yes, and if you shrink the semi circle back to a point charge $Q$ = 3 μC at distance $R$ from the center, the potential will still be the same and you will be back where you started.

Thank you so much for clarifying and sorry it took so long for me to understand the issue

 

[kuruman]

You are welcome. I'm glad your misconception was dispelled and it didn't take as long as you make it sound.