Question

In a source free region in vacuum, if the electrostatic potential $\varphi =4x^2+y^2+c{z}^2$, the value of constant c will be

• A. 2
• B. -3
• C. -5
• D. 4

Answer 1

The correct option is C -5

Given $\varphi =4x^2+y^2+c{z}^2$
In source free region,

$\nabla .\overrightarrow{D}=0$
Also $D=ϵE$
So, $ϵ(\nabla .\overrightarrow{E})=0$

or, $\nabla .\overrightarrow{E}=0$

Also $\overrightarrow{E}=-\nabla V=-\nabla \varphi$

$\nabla V=\frac{\partial \varphi }{\partial x}\widehat{a_x}+\frac{\partial \varphi }{\partial y}\widehat{a_y}+\frac{\partial \varphi }{\partial z}\widehat{a_z}$

$-\overrightarrow{E}=8x\widehat{a_x}+2y\widehat{a_y}+2cz\widehat{a_z}$

Again, $\nabla .\overrightarrow{E}=0$

$\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=8+2+2c=0$

c = - 5