Introductory Physics Volume Two

40 Electric Potential 2.1
so that
∫ B
∆V = −
= −
= −
= −α
= −α
= −α
A
∫ 1
0
∫ 1
0
∫ 1
0
∫ 1
0
∫ 1
0
= −αab 2
⃗E · ⃗dr
⃗E(⃗r(t)) · d⃗r
dt dt
α [ y 2 î + 2xyĵ ] · [aî + bĵ] dt
[ (bt)2î + 2(at)(bt)ĵ ] · [aî + bĵ] dt
[
(bt) 2 a + 2(at)(bt)b ] dt
3ab 2 t 2 dt
In general, to compute a line integral, one must first find a parameterization
of the path.
The following theorem gives a way of finding the electric field if
the electric potential field is already know.
Theorem: Electric Field from the Electric Potential
[
⃗E = −∇V ⃗ ∂V
= −
∂x î + ∂V
∂y ĵ + ∂V ]
∂z ˆk
The symbol ⃗ ∇ represents the gradient operator ∂
Thus the expression ⃗ ∇f is equal to ∂f
∂x î + ∂f
∂y ĵ + ∂f
∂z ˆk.
∂xî + ∂
∂y ĵ + ∂ ∂z ˆk.
Example
The nonuniform electric field, ⃗ E = αy 2 î + 2αxyĵ, that we used in the
previous example, has the electric potential V = −αxy 2 . Let us check
the above theorem.
⃗E = − ⃗ ∇V
= ∂
∂x αxy2 î + ∂ ∂y αxy2 ĵ + ∂ ∂z αxy2ˆk
= αy 2 î + 2αxy ĵ + 0 ˆk OK