Question

The vector field F is shown in the -plane and looks the same in all other horizontal planes. (In other words, F is independent of and its -component is 0.)

(a) Is div Fpositive, negative, or zero? Explain.

(b) Determine whether curl . If not, in which direction does curl F point?

Short answer

(a) The\({\mathop{\rm div}\nolimits} {\bf{F}}\) is \(0\).

(b) The\({\mathop{\rm curl}\nolimits} {\bf{F}}\) is not equal to zero and the value is \( - \frac{{\partial P}}{{\partial y}}{\rm{k}}\). The\({\mathop{\rm curl}\nolimits} {\bf{F}}\) points in negative z direction.

Step-by-step solution

Concept of Divergence of the vector field

Divergenceis avector operatorthat operates on avector field, producing ascalar fieldgiving the quantity of the vector field's source at each point.

\(\begin{array}{l}{\mathop{\rm div}\nolimits} {\bf{F}} = \nabla \cdot {\bf{F}}\\{\mathop{\rm div}\nolimits} {\bf{F}} = \frac{{\partial {F_x}}}{{\partial x}} + \frac{{\partial {F_y}}}{{\partial y}} + \frac{{\partial {F_z}}}{{\partial z}}\end{array}\)

Calculation for the\({\mathop{\rm div}\nolimits} {\bf{F}}\)

(a)

Consider the standard equation of a divergence of vector field for \({\rm{F}} = P{\rm{i}} + Q{\rm{j}} + R{\rm{k}}\) \({\mathop{\rm div}\nolimits} {\rm{F}} = \frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)\)

As \({\bf{F}}\) is independent of \(z\)-component \((R = 0)\) is \(\frac{{\partial R}}{{\partial x}} = \frac{{\partial R}}{{\partial y}} = \frac{{\partial R}}{{\partial z}} = 0\).

We know, the \(y\)-component of each vector is \(0(Q = 0)\), hence \(\frac{{\partial Q}}{{\partial x}} = \frac{{\partial Q}}{{\partial y}} = \frac{{\partial Q}}{{\partial z}} = 0\)and \(P\) decreases as \(y\) increases, so \(\frac{{\partial P}}{{\partial y}} > 0\) but \(P\) doesn't change in the \(x\)-direction or \(z\)-direction, so \(\frac{{\partial P}}{{\partial x}} = \frac{{\partial P}}{{\partial z}} = 0.\)

Substitute \(0\) for \(\frac{{\partial P}}{{\partial x}},\frac{{\partial Q}}{{\partial y}}\) and \(\frac{{\partial R}}{{\partial z}}\) in equation \(\left( 1 \right),\)

\(\begin{array}{l}{\mathop{\rm div}\nolimits} {\rm{F}} = 0 + 0 + 0\\{\mathop{\rm div}\nolimits} {\rm{F}} = 0\end{array}\)

Thus, the\({\mathop{\rm div}\nolimits} {\bf{F}}\) is \(0\).

Calculation for the\({\mathop{\rm curl}\nolimits} {\bf{F}}\)

(b)

Consider the standard equation of a\({\mathop{\rm curl}\nolimits} {\bf{F}}\)for\({\rm{F}} = P{\rm{i}} + Q{\rm{j}} + R{\rm{k}}\). \({\mathop{\rm curl}\nolimits} {\bf{F}} = \left( {\frac{{\partial R}}{{\partial y}} - \frac{{\partial Q}}{{\partial z}}} \right){\rm{i}} + \left( {\frac{{\partial P}}{{\partial z}} - \frac{{\partial R}}{{\partial x}}} \right){\rm{j}} + \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right){\rm{k}}\;\;\;\;\;\;(2)\)

Substitute \(0\) for and \(\frac{{\partial P}}{{\partial y}}\) in equation \(\left( 2 \right),\)

\(\begin{array}{l}{\mathop{\rm curl}\nolimits} {\bf{F}} = (0 - 0){\rm{i}} + (0 - 0){\rm{j}} + \left( {0 - \frac{{\partial P}}{{\partial y}}} \right){\rm{k}}\\{\mathop{\rm curl}\nolimits} {\bf{F}} = (0){\rm{i}} + (0){\rm{j}} + \left( { - \frac{{\partial P}}{{\partial y}}} \right){\rm{k}}\\{\mathop{\rm curl}\nolimits} {\bf{F}} = - \frac{{\partial P}}{{\partial y}}{\rm{k}}\end{array}\)

As \(\frac{{\partial P}}{{\partial y}} > 0\), then the \( - \frac{{\partial P}}{{\partial y}}{\rm{k}}\) is a vector, which is pointing in the negative \(z\)-direction.

Therefore, the \({\mathop{\rm curl}\nolimits} {\bf{F}}\) is not equal to zero and the value is \( - \frac{{\partial P}}{{\partial y}}{\rm{k}}\).