Q111 / 125 · BITSAT 2010 · Physics
In a region, the potential is represented by $\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{x}-8 \mathrm{xy}-8 \mathrm{y}+6 \mathrm{yz},$ where $\mathrm{V}$ is in volts and $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1,1,1) is :
- A$6 \sqrt{5} \mathrm{~N}$
- B$30 \mathrm{~N}$
- C$24 \mathrm{~N}$
- Correct answerD$4 \sqrt{35} \mathrm{~N}$
Solution
$\overrightarrow{\mathrm{E}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} \hat{\mathrm{i}}-\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \hat{\mathrm{j}}-\frac{\partial \mathrm{V}}{\partial \mathrm{z}} \hat{\mathrm{k}}$
$=-[(6-8 \mathrm{y}) \hat{\mathrm{i}}+(-8 \mathrm{x}-8+6 \mathrm{z}) \hat{\mathrm{j}}+(6 \mathrm{y}) \hat{\mathrm{k}}]$
$\mathrm{At}(1,1,1), \overrightarrow{\mathrm{E}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}$
$\Rightarrow \quad(\overrightarrow{\mathrm{E}})=\sqrt{2^{2}+10^{2}+6^{2}}=\sqrt{140}=2 \sqrt{35}$
$\therefore \quad \mathrm{F}=\mathrm{q} \overrightarrow{\mathrm{E}}=2 \times 2 \sqrt{35}=4 \sqrt{35}$
Hence, the correct answer is option: 'D'.