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

Step 1: Recall Gauss's law in differential form.
In free space, the electric flux density is \(\vec{D} = \epsilon_0\vec{E}\), and Gauss's law says \(\nabla\cdot\vec{D} = \rho_v\). So the volume charge density is
\[ \rho_v = \epsilon_0(\nabla\cdot\vec{E}) \]

Step 2: Write out the divergence.
For \(\vec{E} = E_x\hat{i}+E_y\hat{j}+E_z\hat{k}\), the divergence is
\[ \nabla\cdot\vec{E} = \frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z} \]

Step 3: Differentiate each component.
\[ E_x = 2x+5y+6z \ \Rightarrow\ \frac{\partial E_x}{\partial x} = 2 \]
\[ E_y = 5x+4y+10z \ \Rightarrow\ \frac{\partial E_y}{\partial y} = 4 \]
\[ E_z = 6x+10y+2z \ \Rightarrow\ \frac{\partial E_z}{\partial z} = 2 \]

Step 4: Add them up and multiply by \(\epsilon_0\).
\[ \nabla\cdot\vec{E} = 2+4+2 = 8 \]
\[ \rho_v = 8\epsilon_0 \ \text{C/m}^3 \]

Step 5: Check the other options.
Option (B), \(10\epsilon_0\), comes from wrongly adding the coefficient of \(y\) in \(E_x\) or \(E_z\) into the sum instead of only the matching-variable coefficients. Option (C), \(6\epsilon_0\), misses one of the three terms. Option (D), \(4\epsilon_0\), comes from adding only two of the three partial derivatives. Only the coefficient of \(x\) in \(E_x\), of \(y\) in \(E_y\), and of \(z\) in \(E_z\) belong in the divergence, all other cross terms vanish on differentiation.

Final Answer:
\[ \boxed{\rho_v = 8\epsilon_0\ \text{C/m}^3} \]