Field Problems: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

Problems on Field Characteristics

Problem 1

The gravitation field is characterized by a vector of force of gravity per unit of mass of a probe object (a point-mass) as a function of its distance from the source of the field (like Earth).
The direction of this force is always towards a center of mass that is the source of gravity.

Another characteristic of a gravitational field is its potential, which is a work (a scalar) needed to bring a unit of mass from infinity to a point of interest (see "Energy" part of this "Physics 4 Teens" course, chapter "Potential Energy", topic "Gravity").

Let the source of gravitational field be at location {0,0,0} and its mass is M.
Our point of interest is at location {x,y,z}.
The gravitational potential is inversely proportional to a distance from the center of the source of gravity and can be expressed as a scalar function of three coordinates
U(x,y,z) = −G·M·(x²+y²+z²)^−½
where G is a gravitational constant.

Determine the vector
grad U(x,y,z) = V(x,y,z) of gradient of this field at any point {x,y,z}, describe its direction and magnitude.

Solution

Gradient of a scalar field is a vector directed towards the highest rate of change of the value of a scalar field.
The components of a vector of gradient are partial derivatives of a scalar field by each coordinates
grad U(x,y,z) = V(x,y,z) =
= {∂U/∂x,∂F/∂y,∂F/∂z}
which can also be written using a symbol ∇ as
grad U(x,y,z) = V(x,y,z) =
= ∇U(x,y,z)
that looks like a multiplication of a pseudo-vector
∇ = {∂/∂x,∂/∂y,∂/∂z}
by a scalar U(x,y,z).

In our case of
U(x,y,z) = −G·M·(x²+y²+z²)^−½

Partial derivative by argument x of this function is
∂U(x,y,z)/∂x =
= G·M·x·(x²+y²+z²)^−3/2
Similarly,
∂U(x,y,z)/∂y =
= G·M·y·(x²+y²+z²)^−3/2
and
∂U(x,y,z)/∂z =
= G·M·z·(x²+y²+z²)^−3/2

The direction of vector grad U(x,y,z) is along the line from the origin of coordinates towards a location of a point of interest {x,y,z}.

The magnitude of this vector is
|grad U(x,y,z)| =
= G·M·(x²+y²+z²)⁻¹
which happened to be the magnitude of the gravitational force of mass M onto a unit mass (that is, the field intensity) at location {x,y,z}, according to the Newton's Law of Gravitation.
The only difference between the force of gravity and the gradient of a gravitational potential field is, the gradient directed from the center to point {x,y,z}, while the force of gravity is in the opposite direction.


Problem 2

Given a three-dimensional vector field
V(x,y,z) = {0,0,z}

(a) Describe this vector field verbally.
(b) Is there a front surface of this field, where all vectors at it are equal?
(c) Find a divergence of this field.
(d) Is there a part of space where divergence is zero?
(e) Find a curl of this field.

Solution

(a) For any point {x,y,z} the vector V(x,y,z)={0,0,z} has only Z-component not equal to zero. It means, all vectors of this vector field are parallel to Z-axis with zero projection on two other axes.
(b) All vectors at points that belong to the same plane parallel to XY-plane are of the same magnitude and direction since their Z-components are the same, while X- and Y-components are zero. So, any
(c) Divergence of V(x,y,z) is
div V(x,y,z) =
= div (0·i + 0·j + z·k) =
= ∂0/∂x + ∂0/∂y + ∂z/∂z =
= 0 + 0 + 1 = 1
(d) No. The divergence div V(x,y,z) is equal to 1 everywhere.
(e) Curl of V(x,y,z) is
curl V(x,y,z) =
= curl (0·i + 0·j + z·k) =
= (∂z/∂y−∂0/∂z)·i +
+ ∂0/∂z−∂z/∂x)·j +
+ ∂0/∂x−∂0/∂y)·k) =
= (0·i + 0·j + 0·k) = 0

Problem 3

Given a two-dimensional vector field
V(x,y) =
= {−y/√(x²+y²), x/√(x²+y²)}

(a) Describe this vector field verbally.
(b) Is there a front line of this field, where all vectors at it are equal?
(c) Find a divergence of this field.
(d) Is there a part of space where divergence is zero?
(e) Find a curl of this field.

Solution

(a) This vector field is circular, that is a vector V(x,y) at any point of interest is perpendicular to a radius-vector to this point from the origin of coordinates, since the scalar product of V(x,y) and r(x,y)={x,y} is zero.
(b) Yes. Any radial line from the origin of coordinates represents a front line because any field vector originated on it have the same direction (perpendicular to this radial line) and magnitude of 1.
(c) Divergence of V(x,y) is
div V(x,y) =
= div [−y/√(x²+y²)]·i +
+ [x/√(x²+y²)]·j =
= ∂[−y/√(x²+y²)]/∂x +
+ ∂[x/√(x²+y²)]/∂y =
= (−y)·(−x)·(x²+y²)^−3/2 +
+ (x)·(−y)·(x²+y²)^−3/2 =
= (−y·x+x·y)·(x²+y²)^−3/2 = 0
(d) Yes, everywhere the divergence is zero. The equality of the magnitude of the field vectors plays a major role.
(e) Curl of V(x,y) is
curl V(x,y) =
= curl [−y/√(x²+y²)]·i +
+ [x/√(x²+y²)]·j =
= {∂[x/√(x²+y²)]/∂x −
− ∂[−y/√(x²+y²)]/∂y}·k =
= x²·(x²+y²)^−3/2 +
+ y²·(x²+y²)^−3/2 =
= 1/√(x²+y²)
As we see, the curl is the same for all points of interest on the same distance from the origin of coordinates and the curl is diminishing as a point of interest is moving away from the origin of coordinates.