Question

If \( \mathbf{F} = ax \hat{i} + by \hat{j} + cz \hat{k} \), where \( a, b, c \) are constants, then

• A. 0
• B. \( \frac{4}{3} \pi (a + b + c)^2 \)
• C. \( \frac{4}{3} \pi (a + b + c) \)
• D. 1

Hint:

The integral of the force field over the surface of a sphere can be simplified based on symmetry considerations.

Answer 1

The Correct Option is C

We are given the vector field \( \mathbf{F} = ax\hat{i} + by\hat{j} + cz\hat{k} \), where \( a \), \( b \), and \( c \) are constants.

The flux through a surface is given by the surface integral of the dot product of the vector field \( \mathbf{F} \) and the normal vector \( \hat{n} \) of the surface:

\[ \Phi = \iint_S \mathbf{F} \cdot \hat{n} \, dA \]

For the vector field \( \mathbf{F} = ax\hat{i} + by\hat{j} + cz\hat{k} \), the divergence of the field \( \mathbf{F} \) is:

\[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(ax) + \frac{\partial}{\partial y}(by) + \frac{\partial}{\partial z}(cz) = a + b + c \]

Now, the total flux \( \Phi \) over a volume \( V \) is given by the volume integral of the divergence:

\[ \Phi = \iiint_V (\nabla \cdot \mathbf{F}) \, dV = (a + b + c) \iiint_V \, dV \]

The volume of a unit sphere is \( \frac{4}{3} \pi r^3 \), and since we are working with a unit sphere, \( r = 1 \), so the volume is \( \frac{4}{3} \pi \).

Thus, the total flux is:

\[ \Phi = \frac{4}{3} \pi (a + b + c) \]

This corresponds to option (C).