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...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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` .

**Further Reading**