The divergence of \( \mathbf{A} = (1/r^2)\mathbf{a}_r \) in spherical coordinates, away from the origin, is
Answer: A
The spherical divergence is \( (1/r^2)\partial(r^2 A_r)/\partial r \), and with \( A_r = 1/r^2 \) the bracket is a constant so the derivative vanishes. This is why a point charge gives a field with zero divergence everywhere except at the charge itself.
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The curl of \( \mathbf{A} = y\mathbf{a}_x \) is
Answer: B
The z component of the curl is \( \partial A_y/\partial x - \partial A_x/\partial y \), which here is \( 0 - 1 = -1 \). A shear field of this kind does circulate, which is why a single component does not imply a vanishing curl.
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The gradient of \( V = x^2 y \) is
Answer: C
Correct answer: \( 2xy\,\mathbf{a}_x + x^2\mathbf{a}_y \)
Each component of the gradient is the partial derivative with respect to that coordinate: \( \partial V/\partial x = 2xy \), \( \partial V/\partial y = x^2 \), and \( \partial V/\partial z = 0 \). Note that y is held constant while differentiating with respect to x, which is what leaves the y in the first component.
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The Laplacian of \( V = x^2 + y^2 + z^2 \) is
Answer: D
Each term differentiated twice with respect to its own variable gives two, and the three add to six. Since the result is not zero, this potential is a solution of Poisson's equation with a uniform charge density rather than of Laplace's equation.
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The volume of a sphere of radius \( a \) obtained by integrating the spherical volume element is
Answer: A
Integrating \( r^2 \, dr \) from zero to a gives \( a^3/3 \), the sine integral over theta gives two and the phi integral gives \( 2\pi \). The product is \( 4\pi a^3/3 \), which is the standard check on the volume element.
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The scalar triple product of \( \mathbf{a}_x \), \( 2\mathbf{a}_y \) and \( 3\mathbf{a}_z \) is
Answer: B
The three are mutually perpendicular, so the parallelepiped is a rectangular box of sides one, two and three and its volume is six. A vanishing triple product would mean the three vectors were coplanar, which mutually perpendicular vectors are not.
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The flux of \( \mathbf{D} = r\,\mathbf{a}_r \) out of a sphere of radius \( a \) is
Answer: C
On the surface the field has magnitude a and is everywhere normal to it, so the flux is a multiplied by the area \( 4\pi a^2 \). The divergence theorem gives the same answer, since the divergence of the field is three throughout the volume.
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The directional derivative of a scalar V along a unit vector \( \mathbf{a} \) is
Answer: D
It is the projection of the gradient onto the chosen direction, and it is greatest when the direction coincides with the gradient and zero when it is perpendicular. That perpendicular case is movement along a surface of constant V.
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True or false: the unit vectors of the spherical coordinate system are constant vectors.
Answer: A
They have constant magnitude but not constant direction: the radial unit vector points differently at every point. Only the Cartesian unit vectors are genuinely constant, which is why differentiation is so much simpler in that system.
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True or false: a field that is both solenoidal and irrotational must be zero.
Answer: B
A uniform field has zero divergence and zero curl and is plainly not zero. What the two conditions together require is that the field derive from a potential satisfying Laplace's equation, which has many non-trivial solutions.
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