Peoples Physics Book Version-3

12.3. The Coulomb Force Law http://www.ck12.org

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Electric Fields and Electric Forces

Gravity and the Coulomb force have a nice property in common: they can be represented byfields. Fields are a kind
of bookkeeping tool used to keep track of forces. Take the electromagnetic force between two charges given above:

~Fe=kq^1 q^2

r^2

If we are interested in the acceleration of the first charge only — due to the force from the second charge — we can
rewrite this force as the product ofq 1 andkqr 22. The first part of this product only depends on properties of the object
we’re interested in (the first charge), and the second part can be thought of as a property of the point in space where
that object is.

In fact, the quantitykqr 22 captures everything about the electromagnetic force on any object possible at a distancer

fromq 2. If we had replacedq 1 with a different charge,q 3 , we would simply multiplyq 3 bykqr 22 to find the new force

on the new charge. Such a quantity,kqr 22 here, is referred to as the electric field from chargeq 2 at that point: in this
case, it is the electric field due to a single charge:

E~f=kq

r^2

[2] Electric field due to point chargeq,distanceraway

The electric field is a vector quantity, and points in the direction that a force felt by a positive charge at that point
would. If we are given the electric field at some point, it is just a matter of multiplication — as illustrated above —
to find the force any chargeq 0 would feel at that point:

F~e

︸︷︷︸

Force on chargeq 0

= E~f

︸︷︷︸

Field

×q 0

︸︷︷︸

Charge

Force on chargeq 0 in an electric field

Note that this is true forallelectric fields, not just those from point charges. In general, theelectric fieldat a point
is the force apositive test charge of magnitude 1would feel at that point. Any other charge will feel a force along
the same line (but possibly in the other direction) in proportion to its magnitude. In other words, the electric field
can be though of as “force per unit charge”.

In the case given above, the field was due to a single charge. Such a field is shown in the figure below. Notice that
this a field due to a positive charge, since the field arrows are pointing outward. The field produced by a point charge
will be radially symmetric i.e., the strength of the field only depends on the distance,r, from the charge, not the
direction; the lengths of the arrows represent the strength of the field.