Consider the vector field

**E**(x,y,z) = a(x^{2} - y^{2} + z^{2}, z^{2}
- 2xy, 2zy + 2zx),

where a is a constant expressed in the appropriate units.

(a) Is this field irrotational?

(b) What is the corresponding charge density?

Solution:

- Concepts:

Maxwell's equations - Reasoning:

**∇**×**E**= 0? If yes, we have a static field.**∇**∙**E**= ρ/ε_{0}. - Details of the calculation:

**∇**×**E**= (∂E_{z}/∂y - ∂Ey/∂z)**i**+ (∂E_{x}/∂z - ∂E_{z}/∂x)**j**+ (∂E_{y}/∂x - ∂E_{x}/∂y)**k**∂E

_{z}/∂y -∂E_{y}/∂z = a(2z - 2z) = 0, ∂E_{x}/∂z - ∂E_{z}/∂x = a(2z - 2z) = 0,

∂E_{y}/∂x - ∂Ex/∂y = a(-2y + 2y) = 0.

**∇**×**E**= 0.

The field is irrotational.

(b)**∇**∙**E**= ρ/ε_{0},

**∇**∙**E**= ∂E_{x}/∂x + ∂Ey/∂y + ∂E_{z}/∂z = a(2x - 2x + 2y + 2x) = 2a(x + y)

ρ = 2aε_{0}(x + y)

In a volume of space V the electric field is

**E** = c(2x^{2} -2xy - 2y^{2})**i** + c(y^{2}
- 4xy -x^{2})**j**,

where c is a constant.

(a) Verify that this field represents an electrostatic field.

(b) Determine the charge density ρ in the volume V consistent with this field.

Solution:

- Concepts:

Maxwell's equations, conservative fields - Reasoning:

Conservative electrostatic fields are irrotational,**∇**×**E**= 0.

Details of the calculation:

(**∇**×**E**)_{x}_{z}/∂y - ∂E_{y}/∂z = 0, (**∇**×**E**)_{y}_{x}/∂z - ∂E_{z}/∂x = 0.

**E**has no z-component and the x- and y-components do not depend on z.

**(∇**×**E**)_{z}_{y}/∂x - ∂E_{x}/∂y = c(-4y - 4y) - c(-2x - 4y) = 0.

**E**is irrotational and therefore an electrostatic, conservative field.

(b) ρ/ε_{0}=**∇·E**= ∂E_{x}/∂x + ∂E_{y}/∂y = c(4x - 2y) + c(2y - 4x) = 0.

Charges outside the volume V or on its surface must produce this field.

Can the following vector functions represent static
electric fields? If yes, determine the charge density. This is not a yes/no
question. You must justify your answer mathematically.

(a)** E** = **r**×(**c**×**r**) (Here
**c** is a constant vector.)

(b)** E** = c r **r** (Here c is a constant and r = |**r**|.)

Solution:

- Concepts:

Maxwell's equations - Reasoning:

In electrostatics**∇·E**= ρ/ε_{0},**∇×E**= 0, (SI units). - Details of the calculation:

(a)**E**=**r**×(**c**×**r**) =**c**(**r·r**) -**r**(**c·r**) =**c**r^{2}-**r**(**c·r**).

**∇**×**E**=**∇**×(**c**r^{2}-**r**(**c·r**)) =**∇**r^{2}×**c**-**∇**×**r**(**c·r**) = 2r**∇**r×**c**-**∇**×**r**(**c·r**).

**∇**r =**r**/r.**∇**×**r**(**c·r**) = (**c·r**)**∇**×**r**-**r**×**∇**(**c·r**)**= ∇**(**c·r**)×**r**since**∇**×**r**= 0.

**∇**(**c·r**) =**i**c_{x}+**j**c_{y}**k**c_{z}=**c.**×

∇**E**= 2**r**×**c**-**c**×**r**= 3**r**×**c**≠ 0.

**E**=**r**×(**c**×**r**) cannot represent a static electric field.(b)

**E**= c r**r**,**∇**×**E**= (**∇·**cr**)**×**r**+ cr**∇**×**r**= (c/r)**r**×**r =**0.

**E**= c r**r**can represent a static field.

**∇·E**=**∇·**c r**r**= c**∇**r**·r**+ cr**∇·r**= (c/r)**r·r**+ 3cr = cr + 3cr = 4cr = ρ/ε_{0}.

ρ = ε_{0}4cr.

The electric field in some volume V of space is
**E** = [2A/(x^{3}y^{2})
- By]**i** + [2A/(x^{2}y^{3}) - Bx]**j**.

(a) Show that the line integral of this field over a close path is zero.

(b) Which charge density ρ in the volume V is consistent with this field?