Question

An electric field in free space is
\[ \vec{E} = (2x+5y+6z)\hat{i} + (5x+4y+10z)\hat{j} + (6x+10y+2z)\hat{k} \ \text{V/m} \]
The charge density is \(\_\_\_\_\) \(\text{C/m}^3\). (\(\epsilon_0\) is the permittivity of free space)

• A. \(8\epsilon_0\)
• B. \(10\epsilon_0\)
• C. \(6\epsilon_0\)
• D. \(4\epsilon_0\)

Hint:

Use \(\rho_v=\epsilon_0(\nabla\cdot\vec{E})\), and differentiate each component of \(\vec{E}\) only with respect to its own matching variable.

Answer 1

The Correct Option is A

This question checks the differential form of Gauss's law, linking the divergence of the electric field directly to the volume charge density: \(\rho_v = \epsilon_0(\nabla\cdot\vec{E})\).

Since every component of \(\vec{E}\) is a simple linear function of \(x\), \(y\), and \(z\), we can shortcut the differentiation: the partial derivative of a linear term with respect to its own matching variable is just the coefficient of that variable, and every other term that does not contain that variable drops to zero.

1. The \(x\)-component is \(2x+5y+6z\). Its coefficient of \(x\) is 2, so \(\dfrac{\partial E_x}{\partial x} = 2\).
2. The \(y\)-component is \(5x+4y+10z\). Its coefficient of \(y\) is 4, so \(\dfrac{\partial E_y}{\partial y} = 4\).
3. The \(z\)-component is \(6x+10y+2z\). Its coefficient of \(z\) is 2, so \(\dfrac{\partial E_z}{\partial z} = 2\).

Adding these three coefficients gives the divergence directly, without writing out the full derivative each time:

$$ \nabla\cdot\vec{E} = 2+4+2 = 8 $$

So the charge density is

$$ \rho_v = \epsilon_0\times8 = 8\epsilon_0\ \text{C/m}^3 $$

Let's summarize:

• Only the coefficient of the matching variable in each field component contributes to the divergence.
• The divergence works out to 8, so the charge density is \(8\epsilon_0\), matching option (A).

$$ \boxed{8\epsilon_0} $$