Electric potential due to multiple charges

[hasankamal007]

Hello!

For Single Charge:
I studied the formula for electric potential for a point charge (V=kq/r). It was derived when the test charge approached the source charge from infinity "head-on". In this case the electric field due to source charge and displacement of test charge were vectors in same(or at 180degrees) direction. Fine.

For Multiple Charge:
Now, suppose there's a cube with +q point charges at its vertices. Now, to calculate the potential, let's say at the centre of the cube we add potentials due to all those eight charges using v=kq/r where we put r as the distance between vertice and centre of cube.

But this should work only if the test charge brought from infinity is head-on to every charge at vertice of cube! And you cannot do that since in this arrangement, if you try to bring the test charge from infinity head on to anyone charge out of eight, you will start having an angle between other six charges.

Hence, we must take account of cos theta and stuff because then electric field (of source) and displacement (of test charge) are at an angle to each other. So why do we straightaway use that formula?

 

[Chestermiller]

Hello!

For Single Charge:
I studied the formula for electric potential for a point charge (V=kq/r). It was derived when the test charge approached the source charge from infinity "head-on". In this case the electric field due to source charge and displacement of test charge were vectors in same(or at 180degrees) direction. Fine.

For Multiple Charge:
Now, suppose there's a cube with +q point charges at its vertices. Now, to calculate the potential, let's say at the centre of the cube we add potentials due to all those eight charges using v=kq/r where we put r as the distance between vertice and centre of cube.

But this should work only if the test charge brought from infinity is head-on to every charge at vertice of cube! And you cannot do that since in this arrangement, if you try to bring the test charge from infinity head on to anyone charge out of eight, you will start having an angle between other six charges.

Hence, we must take account of cos theta and stuff because then electric field (of source) and displacement (of test charge) are at an angle to each other. So why do we straightaway use that formula?

For anyone of the charges, the electric field is radially (spherically) symmetric. So any non-radial components of differential displacement, when dotted with the electric field vector are zero. So only the radial component of differential displacement contributes to the work required to move a test charge from infinity to any arbitrary location. This applies to each and every charge of the multicharge array.

Chet

 

[hasankamal007]

For anyone of the charges, the electric field is radially (spherically) symmetric. So any non-radial components of differential displacement, when dotted with the electric field vector are zero. So only the radial component of differential displacement contributes to the work required to move a test charge from infinity to any arbitrary location. This applies to each and every charge of the multicharge array.

Chet

Exactly Chet!,
the dot product should zero out the perpendicular component of electric field on the test charge(W=F.s.cosθ).
So, we shouldn't use V=kq/r for a such-arranged multi charged system because this equation was itself derived only for head-on cases.
There should be something like cosθ involved somewhere due to the dot product. But still, only (kq/r+kq/r+...) is used everywhere regardless of this problem in such-arranged(cubic) multi charge system. So, what to do?

For instance, I was solving this question in my physics assignment I described earlier(charges on cube's vertex) AND you get correct answer if you use V=kq/r+kq/r...=8kq/r.
But now we're saying V=kq/r isn't valid here. Then, who's right?

 

[Chestermiller]

Exactly Chet!,
the dot product should zero out the perpendicular component of electric field on the test charge(W=F.s.cosθ).
So, we shouldn't use V=kq/r for a such-arranged multi charged system because this equation was itself derived only for head-on cases.
There should be something like cosθ involved somewhere due to the dot product. But still, only (kq/r+kq/r+...) is used everywhere regardless of this problem in such-arranged(cubic) multi charge system. So, what to do?

For instance, I was solving this question in my physics assignment I described earlier(charges on cube's vertex) AND you get correct answer if you use V=kq/r+kq/r...=8kq/r.
But now we're saying V=kq/r isn't valid here. Then, who's right?

The combined electric field from the several charges is equal to the linear superposition of the electric fields from the individual charges. So all we really need to be able to prove is that the work to bring a test charge from infinity to any arbitrary location relative to a single charge q is independent of the path we take. Now, if we put the single charge q at the origin, the electric field is:
$$\vec{E}=\frac{kq}{r^2}\vec{i}_r$$
Let $\vec{ds}$ be an arbitrary displacement of the test charge. Then,
$$\vec{ds}=(ds)_r\vec{i}_r+(ds)_θ\vec{i}_θ+(ds)_{\phi}\vec{i}_{\phi}$$
Now, in spherical coordinates, (ds)_r = dr, so:
$$\vec{ds}=\vec{i}_rdr+(ds)_θ\vec{i}_θ+(ds)_{\phi}\vec{i}_{\phi}$$
If we take the dot product of this arbitrary displacement vector with the electric field vector, we get:
$$(\vec{E}\centerdot \vec{ds})=\frac{kq}{r^2}dr$$
Note that there are no contributions from the components of the displacements in the other two coordinate directions. So, the work to bring a test charge from infinity to any arbitrary location relative to the charge is independent of path.

Chet

 

[sardar]

I understand your confusion about using the formula for electric potential for a point charge (V=kq/r) in the case of multiple charges. However, it is important to note that the formula for electric potential is derived from the fundamental laws of electrostatics, including Coulomb's law and the principle of superposition. This means that the electric potential at any point is the sum of the potentials due to each individual charge.

In the case of multiple charges, the electric field and displacement of the test charge may not always be in the same direction, as you mentioned. However, the principle of superposition states that the total electric field at any point is the vector sum of the individual electric fields due to each charge. This means that we can still use the formula for electric potential for a point charge, but we must take into account the direction of the electric fields.

In other words, the formula for electric potential takes into account the direction of the electric field and displacement, even in the case of multiple charges. This is why we do not need to use cos theta and other factors in the formula.

Furthermore, it is important to note that the formula for electric potential is an approximation and may not be exact in all cases. As scientists, we must always consider the limitations and assumptions of our models and formulas, and adjust accordingly when necessary.

I hope this helps to clarify your confusion about using the formula for electric potential in the case of multiple charges. Remember, as scientists, we must always question and analyze our understanding of the natural world in order to advance our knowledge and understanding.