Example 7-14. Let r =xˆe 1 +yˆe 2 +zˆe 3 denote the position vector to a general
point (x, y, z ). Show that
grad(^1
r) = grad^1
|r |=−^1
r^2grad (r) = −^1
r^3r =−^1
r^2ˆerwhere ˆeris a unit vector in the direction of r.
Solution Let r=|r |=
√x^2 +y^2 +z^2 so that^1 r= (x^2 +y^2 +z^2 )−^1 /^2. By definition
grad(1
r) =∂
∂x(
1
r)
ˆe 1 + ∂
∂y(
1
r)
eˆ 2 + ∂
∂z(
1
r)
ˆe 3where
∂
∂x(
1
r)
=−^1
2(x^2 +y^2 +z^2 )−^3 /^2 (2 x) = −x
r^3
∂
∂y(
1
r)
=−^1
2(x^2 +y^2 +z^2 )−^3 /^2 (2 y) = −y
r^3
∂
∂z(
1
r)
=−1
2 (x(^2) +y (^2) +z (^2) )− 3 / (^2) (2 z) = −z
r^3
Substituting for the partial derivatives in the gradient gives
grad(1
r) = grad1
|r |=−r
r^3 =−1
r^2(
r
r)
==1
r^2 grad r=−1
r^2ˆerExample 7-15. Let r =xˆe 1 +yˆe 2 +zˆe 3 denote the position vector to a general
point (x, y, z )and let r=|r |. Find grad(rn).
Solution By definition
grad(rn) = ∇(rn) = ∂rn
∂xˆe 1 +∂rn
∂yˆe 2 +∂rn
∂zˆe 3where
∂r n
∂x =nrn− 1 ∂r
∂x =nrn− 1 x
r=rn− (^2) x
∂r n
∂y
=nr n−^1 ∂r
∂y
=nrn−^1 y
r
=nr n−^2 y
∂r n
∂z =r
n− 1 ∂r
∂z =nr
n− 1 z
r=nr
n− (^2) z
so that
grad(rn) = nr n−^2 r =nrn−^1 ˆer