# The electric potential of a very large isolated flat metal plate is V0. It carries a uniform distribution of charge of surface density σ (C/m2), or σ/2 on each surface. Determine V at a distance x from the plate. Consider the point x to be far from the edges and assumex is much smaller than the plate dimensions. Express your answer in terms of the variables V0, σ, x, and appropriate constants. The electric field of a infinite flat plane is uniform and perpendicular to the plane (please see attached reference link.) It's magnitude equals

`E = sigma/(2epsilon_0)` , where `sigma` is the surface charge density on the plane and `epsilon_0` is the dielectric constant.

In this example, since the plate is very large and we...

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The electric field of a infinite flat plane is uniform and perpendicular to the plane (please see attached reference link.) It's magnitude equals

`E = sigma/(2epsilon_0)` , where `sigma` is the surface charge density on the plane and `epsilon_0` is the dielectric constant.

In this example, since the plate is very large and we are considering the distance x very close to the sheet (x is much smaller than the sheet dimensions), we can assume that the field of the plate will be the same as that of an infinite plane.

The electric potential and electric field are related as

`Ex = -(dV)/(dx)` . In this case, the only component of the field is the x-component. So the potential an be found by integrating the field:

`V(x) - V(0) = -int_0 ^ x Edx`

Here, V(x) is the potential the point which is the distance x away from the plate and V(0) is the potential on the plate, which is V_0. Since E is uniform, that is, it does not depend on x, the integral will be

`V(x) -V_0 = -Ex = -sigma/(2epsilon_0) x`

So the potential at point x away from the plate is

`V(x) = V_0 - sigma/(2epsilon_0) x` .