BELIEVE   ME   NOT!    -     A   SKEPTIC's   GUIDE  

The Electric Field

The ELECTRIC FIELD $\Vec{E}$ at any point in space is defined to be the force per unit test charge due to all the other charges in the universe. That is, there is probably no "test charge" q there to experience any force, but if there were it would experience a force

$$\Vec{F}_E \; = \; q \; \Vec{E}$$ (17.5)

Note that since the force is a vector, $\Vec{E}$ is a vector field.

Since by definition $\Vec{E}$ is there even if there isn't any test charge present, it follows that there is an electric field at every point in space, all the time! [It might be pretty close to zero, but it's still there!]^17.4

Is the ELECTRIC FIELD real? No. Yes. You decide.^17.5 This paradigm makes everything so much easier that most Physicists can't imagine thinking about ${\cal E}$&${\cal M}$ any other way. Does this blind us to other possibilities? Undoubtedly.

A single isolated electric "source" charge Q [I am labelling it differently from my "test" charge q just to avoid confusion. Probably it won't work.] generates a spherically symmetric electric field

$$\Vec{E} \; = \; k_E \; {Q \over r^2} \; \hat{r}$$ (17.6)

at any point in space specified by the vector distance $\Vec{r}$ from Q to that point. That is, the field $\Vec{E}$ is radial [in the direction of the radius vector] and has the same magnitude E at all points on an imaginary spherical surface a distance r from Q.

It might be helpful to picture the acceleration of gravity as a similar vector field:

$$\Vec{g} \; = \; -G \; {M_E \over r^2} \; \hat{r}$$ (17.7)

-- i.e. $\Vec{g}$ always points back toward the centre of the Earth (mass M_E) and drops off as the inverse square of the distance r from the centre of the Earth.