5 charges placed at 5 vertices of a regular hexagon

[Apashanka]

Homework Statement

If 5 charges (each q) are placed at 5 vertex of a regular hexagon of side a then effectively the electric field at the centre of the hexagon is $$\frac{q}{4\pi\epsilon_0} $$

Relevant Equations

If 5 charges (each q) are placed at 5 vertex of a regular hexagon of side a then effectively the electric field at the centre of the hexagon is $$\frac{q}{4\pi\epsilon_0} $$

If 5 charges (each q) are placed at 5 vertex of a regular hexagon of side a then effectively the electric field at the centre of the hexagon is $$\frac{q}{4\pi\epsilon_0a^2} $$ but the potential is $$\frac{5q}{4\pi\epsilon_0a}$$ but then what about $$V=-\int \textbf{E•dr}$$

 

[gneill]

If 5 charges (each q) are placed at 5 vertex of a regular hexagon of side a then effectively the electric field at the centre of the hexagon is

A hexagon has six vertexes. Perhaps you meant a pentagon? Or should one vertex be left without a charge?

 

[Apashanka]

A hexagon has six vertexes. Perhaps you meant a pentagon? Or should one vertex be left without a charge?

One vertex left without charge

 

[gneill]

but then what about $$V=-\int \textbf{E•dr}$$

What about it? It's a line integral (or more generally a path integral), so you need to define two endpoints and a path between them.

 

[kuruman]

If you are asking how to find the electric field from the potential, you need to find the potential at a point $\{x,y\}$ off the center, find the gradient $\text{-}\vec \nabla V$ and then calculate it at the origin. However, there is a quicker way to answer this question that exploits symmetry and superposition.

 

[hutchphd]

The electric field is a vector quantity so you need also to specify the direction of the field.

 

[Apashanka]

If you are asking how to find the electric field from the potential, you need to find the potential at a point $\{x,y\}$ off the center, find the gradient $\text{-}\vec \nabla V$ and then calculate it at the origin. However, there is a quicker way to answer this question that exploits symmetry and superposition.

From symmetry the field at the centre due to 4 point charges gets canceled and therefore the field due to all the 5 point charges will actually be due to a single point charge at the centre which is $$\frac{q}{4\pi\epsilon_0a^2}$$ and similarly the potential at the centre due to all the 5 point charges is $$\frac{5q}{4\pi\epsilon_0a}$$
My question is from $$V=-\int E • dr$$ here in this case we get the potential as $$\frac{q}{4\pi\epsilon_0a}$$ which is not consistent with above??

 

[mjc123]

How do you get that potential from that integral? Are you trying to integrate kq/a² with a as the variable? That is not correct. kq/a² is the value of the field at the centre; it is not a formula for the field over a space defined by the supposed variable a.