Solutions

  1. \( \mathrm{div}(\vec{E}) =\vec{\nabla}\circ\vec{E} = ye^{xy} + x\cos(xy) - 2x \cos(zx)\sin(zx) \).
  2. \( \mathrm{div}(\vec{F}) =\vec{\nabla}\circ\vec{F} = 3 \).
  3. Each component \( B_x \), \( B_y \) and \( B_z \) does not depend on \( x \), \( y \) and \( z \) respectively. So straight away we can write \( \frac{\partial B_x}{\partial x} = 0, \frac{\partial B_y}{\partial y} = 0 \) and \( \frac{\partial B_z}{\partial z} = 0 \) therefore \( \mathrm{div}\vec{B} = 0 \).
    1. Zero divergence. \( x \) component does not change if you move in the \(x \) direction i.e. \( \frac{\partial E_x}{\partial x} = 0 \). The same for the \( y \) component. It stays a constant 0 when you move in \( y \) direction i.e. \( \frac{\partial E_y}{\partial y} = 0 \). The \( x \) component does change throughout the field, however it changes w.r.t. \( y \) only!
    2. Zero divergence by the same argument.
    3. Zero divergence by the same argument.
    4. Zero divergence as \( x \) and \( y \) components are constant.
    5. A non zero divergence. The \( x \) component changes as you move in the \( x \) direction i.e. \( \frac{\partial E_x}{\partial x} \neq 0 \). From a non-mathematical point of view, it looks like the vectors are getting larger when you move along the \( x \) axis, so it appears the field is diverging from each point.
    6. A non zero divergence by the same argument. Even though the vectors here have a magnitude of 0, the rate of change is still non-zero.
    7. A non zero divergence by the same argument. Even though vectors point in the opposite direction, the rate of change of these field lines w.r.t. \( x \) is the same, they are becoming more positive the more we more right.
  5. Using Gauss' Law in differential form, \( \vec{\nabla} \circ \vec{E} = 3E_0(x^2 + y^2 + z^3) \) therefore \( \rho(x,y,z) = 3 E_0 \varepsilon_0(x^2 + y^2 + z^2) \).