Work done moving two charges together

[phosgene]

Homework Statement

Two point charges of magnitude +10 μC each are placed 0.2 m away from each other.

a) How much work is done in placing the second charge?

b) Is there any point at which the electric field and electric potential are both 0?

Homework Equations

Work = (charge of first point charge)(ΔV of second charge)

E=kQ/(d^2), where E = electric field, k = proportionality constant, Q = charge of point charge, d = distance from point charge

V=kQ/r, where V = voltage, Q = charge and r = distance from charge

V=Ed, where V = voltage, E = electric field and d = distance from point charge

The Attempt at a Solution

a) I assumed that the point charge was being moved from a spot far enough so that it could be approximated by being moved there from an effectively infinite distance away. Then the work done can be determined as follows:

Work = 10 μC(kQ/r2 - kQ/r1), where r2 = 0.2m and r1 = infinity

These simplifies to Work = (10 μC)(k10μC)/0.2m

b) I said that at a point very far away the electric field and electric potential will be 0. This is because E=kQ/(r^2) at a very far away point will effectively be E=kQ/(infinity), which is approximately 0. As the potential difference is equal to the electric field multiplied by the distance, it too will effectively be 0 at that same far away point.

Is my reasoning correct?? I don't really get voltage, potential difference and electrical potential energy.

EDIT: Sorry! I forgot to put a title! It should be 'Work done moving two charges together and possible point of zero electric field and electric potential'. But I can't edit a title in.

 

[tiny-tim]

hi phosgene!

a) I assumed that the point charge was being moved from a spot far enough so that it could be approximated by being moved there from an effectively infinite distance away. Then the work done can be determined as follows:

Work = 10 μC(kQ/r2 - kQ/r1), where r2 = 0.2m and r1 = infinity

These simplifies to Work = (10 μC)(k10μC)/0.2m

ok (but you need it in joules)

b) I said that at a point very far away the electric field and electric potential will be 0. This is because E=kQ/(r^2) at a very far away point will effectively be E=kQ/(infinity), which is approximately 0. As the potential difference is equal to the electric field multiplied by the distance, it too will effectively be 0 at that same far away point.

i'm not sure that we can talk about a point actually being at infinity

perhaps it would be easier to make use of the fact that both the potential (as a scalar) and the field (as a vector) are additive?

 

[phosgene]

i'm not sure that we can talk about a point actually being at infinity

perhaps it would be easier to make use of the fact that both the potential (as a scalar) and the field (as a vector) are additive?

Thanks for the reply. I don't quite understand what you mean:(. But I think I have a better answer for b). The point exactly in-between the charges has an electric field of zero because the electric field vectors from both charges are exactly equal and opposite. The electric potential at this point is also zero because if a test charge is placed there, it will not move because the forces on it are equal.

 

[tiny-tim]

hi phosgene!

The electric potential at this point is also zero because if a test charge is placed there, it will not move because the forces on it are equal.

no …

zero potential means that it is at the same potential as at infinity

(ie that no net work would be done moving it there from infinity)

compare this with a ball at the top of a mountain …

it won't move, but its gravitational potential is a lot more than at the bottom of the mountain

(ie a lot of work would be done moving it there)

oh, and you can just add potentials​

 

[asteriatic]

Your reasoning is correct. When calculating the work done in placing the second charge, it is important to consider the initial position of the charge and the final position of the charge. In this case, the first charge is being moved from infinity to a distance of 0.2m away from the second charge. Therefore, the work done is equal to the potential energy gained by the second charge, which is calculated using the equation V=kQ/r. This simplifies to Work = (10 μC)(k10μC)/0.2m.

Regarding the point of zero electric field and electric potential, your reasoning is also correct. At a very far away point, the electric field and electric potential will both be approximately zero, as the distance from the charges becomes significantly larger compared to the distance between the charges. This is because the electric field and potential decrease with distance according to the inverse square law. It is important to note that the concept of voltage and potential difference can be difficult to understand, but essentially voltage is a measure of the electric potential energy per unit charge at a specific point in an electric field. Therefore, at a point where the electric field is zero, the voltage will also be zero. I hope this helps clarify your understanding of these concepts.