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lec2.ppt
1. NPTEL β Physics β Mathematical Physics - 1
Lecture 2
Physical examples of gradient
1. An electric dipole moment πβ located at origin, creates a potential at πβ given
by,
π.πβ
V(πβ) = K , where K is a constant.
π3
Find the electric field πΈββ = - βββ V (πβ). Where βββ is an operator given by βββ= ( iΛ π
+ Λj π
+ kΛ π
)
ππ₯ ππ¦ ππ§
Solution
πΈββ = βπ (π₯Μ
π
+ π¦Μ
π
+
π
)
ππ₯ ππ¦ ππ§
(π₯ + π¦ + π§)
(π₯2 + π¦2 + π§2) β2
3
The π₯Μ component yields,
= βπ [
1
(π₯2 + π¦2 + π§2) β2 (π₯2 + π¦2 + π§2) β2
3 5
β
3π₯2
] xΜ
Putting the contribution from yΜ and zΜ
πΈββ =
3π
(π₯2 xΛ + π¦2 yΛ + π§2 zΛ )
β
π rΛ
π5 π3
2. For a position vector πβ, calculate βββππ and use this result to calculate
βββ( ).
1
π
Solution
βββππ = ( π₯Μ π
+
yΛ ππ₯
π
ππ¦ ππ§
+ zΜ
π
)ππ
= π₯Μ πππβ1 ππ
+ yΛ πππβ1 ππ
+ π§Μ πππβ1 ππ
ππ₯ ππ¦ ππ§
= πππβ1[ xΛ ππ
+
yΛ ππ₯
ππ
+ zΜ ππ
]
ππ¦ ππ§
Using π2 = π₯2 + π¦2 + π§2
ππ π₯ ππ
ππ₯ π ππ¦ π ππ§ π
= , = , =
π¦ ππ π§
βββππ = πππβ2 (π₯ xΛ + π¦ yΛ + π§
zΛ )
= πππβ2πβ
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2. NPTEL β Physics β Mathematical Physics - 1
To calculate β ( ), put n =-1
β
β
1
π
βββ
(
1 π
) =
πβ
π3
The potential due to a point change goes as , so electric field goes as
1
π
πβ
π3
Exercise
Two identical point charges of magnitude Q at (1, 0, 0) and (-1, 0, 0). The potential in the xy
plane is given by,
V(x, y, z =0) = [ +
π 1 1
4ππ0 β(π₯β1)2+π¦2 β(π₯+1)2+π¦2 ]
Find the electric field using, πΈββ = - π»ββπ.
Formal definition and properties of a Gradient
Thus gradient represents a direction that is perpendicular to the surface π(πβ). βββπ(πβ) = 0
denotes a minimum or a maximum or a saddle point (saddle point is defined as the one that is minimum in
one direction, but is maximum in the other direction.
Let us consider a scalar function π(π₯, π¦, π§). A change in π is given by,
ππ = (ππ
) ππ₯ + (ππ
) ππ¦ + (ππ
) ππ§ β
ππ¦ πy πz
Example
Find βββ π and |βββ π| for π = 2π₯π§4- π₯2y at the point (2,-2,-1). Solution
π = 2π₯π§4 - π₯2π¦
βββπ = xΛ [2π§4 β 2π₯π¦] + yΛ [βπ₯2] + zΛ [8π₯π§3]
|βββββ π|(2,β2,β1) = 10 xΛ β4π¦ β 16π§3
Then |βββπ| = 2β93
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3. NPTEL β Physics β Mathematical Physics - 1
Exercise problem
The above equation can be written as,
ππ = ββ π. ππ with ββ π = xΛ ππ
+ yΛ ππ
+ zΛ ππ
and ππ is a differential length
element
ππ₯ ππ¦ ππ§
given by,
ππ = π₯Μ ππ₯ + π¦Μ ππ¦ + π§Μ ππ§
The ββ π is a vector where magnitude is the maximum value of ππ
and whose direction is
the ππ
direction in which is maximum.
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ππ
ππ
Exercise
Find a unit vector perpendicular to the surface π₯2 + π¦2 + π§2 = 1 at the point (4,2,3).
Before we wind up our discussion on gradient, we wish to remind the reader that gradient of a function π,
that is βββπ is perpendicular to the surface π = constant. Thus, considering a physical situation that four
sound boxes are located at four corners of the room, each emitting a music of different frequency and
intensity, if we ask a question that what is the distribution of sound intensity in the room. Gradient of this
intensity function will tell you the direction along which the change is most rapid.