184_notes:examples:week2_electric_field_negative_point

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Suppose we have a negative charge with charge Q. What is the magnitude of the electric field at a point P, which is a distance R from the charge? Draw the electric field vector on a diagram to show the direction of the electric field at P. Negative Point Charge -Q, and Point P

Facts

  • The point P is a distance R away from the point charge.
  • The separation vector r points from Q to P, and has magnitude R.

Lacking

  • Magnitude of the electric field at P.
  • Direction of the electric field at P.

Approximations & Assumptions

  • The charge with value Q is a point charge.
  • The electric field at P is made up entirely of contributions from the point charge (there are no other surrounding charges).

Representations

  • The electric field from the point charge can be written as E=14πϵ0qr2ˆr,
    where q represents our charge (Q in this case), r is our distance (R), and ˆr is the unit vector pointing from the point charge to P.
  • We can represent the electric field in our diagram with an arrow, since the electric field at a specific point is a vector.

The electric field at P is given by: E=14πϵ0qr2ˆr

We can plug in our charge (Q) and the magnitude of the separation vector (R) to get: E=14πϵ0(Q)R2ˆr

s-direction drawn in

This leaves us to find the direction of ˆr. The first thing to do would be to draw in the vector rQP. This vector points from the charge -Q to Point P since P is where we want to find the electric field. Now we need to define a coordinate axis. We could pick the normal x- and y-axes, but this would make writing the r and ˆr more difficult because there would be both x and y components to the r-vector (since it points in some diagonal direction).

Instead, let's pick a coordinate direction that falls along the same axis as the rQP. Since this is a coordinate direction that we're naming, let's call this the ˆs direction. That means that rQP points in the ˆs direction, so ˆr=ˆs. Plugging this into our electric field equation gives: E=14πϵ0QR2ˆs

Since the charge is negative, this means that the electric field points in the opposite direction of the r. To make this explicit, we could write this as: E=14πϵ0QR2(ˆs)

This gives the magnitude of the electric field as |E|=14πϵ0QR2

with the direction is given by ˆs, which is the opposite of the direction of rQP. The electric field at P therefore points from P to the point charge. You'll find this to be true for all negative charge - the electric field points towards negative charges. A diagram is shown below. Charge Distribution Induced From Two Sides, Solution

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  • Last modified: 2017/08/28 20:34
  • by tallpaul