Question

In a region, the potential is represented by V(x, y, z) = 6x - 8xy - 8y + 6yz, where $V$ is in volts and $x,y,z$ are in metres. The electric force experienced by a charge of $2$ coulomb situated at point $(1,1,1)$ is

• A. $6\sqrt{5}N$
• B. 30 N
• C. 24 N
• D. $4\sqrt{3}5N$

Answer 1

Answer: (D)

Here, V(x, y, z) = 6x - 8xy - 8y + 6yz
The x, y and z components of electric field are
$E_x=-\frac{\partial V}{\partial x}=-\frac{\partial }{\partial x}(6x-8xy-8y+6yz)$
$=-(6-8y)=-6+8y$
$E_y=-\frac{\partial V}{\partial y}=-\frac{\partial }{\partial y}(6x-8xy-8y+6yz)$
$=-(-8x-8+6z)=8x+8-6z$
$E_z=-\frac{\partial V}{\partial z}=-\frac{\partial }{\partial z}(6x-8xy-8y+6yz)=-6y$
$\overrightarrow{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}$
$=(-6+8y)\hat{i}+(8x+8-6z)\hat{j}-6y\hat{k}$
At point (1, 1, 1)
$\overrightarrow{E}=(-6+8)\hat{i}+(8+8-6)\hat{j}-6\hat{k}$
$=2\hat{i}+10\hat{j}-6\hat{k}$
The magnitude of electric field $\overrightarrow{E}$ is
$\overrightarrow{E}=\sqrt{E_x^2+E_y^2+E_z^2}=\sqrt{(2{)}^2+(10{)}^2+(-6{)}^2}$
$=\sqrt{1}40=2\sqrt{3}5N{C}^{-1}$
Electric force experienced by the charge
$F=qE=2C\times 2\sqrt{3}5N{C}^{-1}=4\sqrt{3}5N$