ALIGN} & {{E}_{x}}\frac{\delta V}{\delta x}=6-8y \\ & {{E}_{y}}=\frac{\delta V}{\delta y}=-8x-8+6z \\ \end{align} \RIGHT\}\] (ii) \[{{E}_{z}}=6y\] Above values of \[{{E}_{x}},{{E}_{y}}\], and \[{{E}_{z}}\] at \[(1,\text{ }1,\text{ }1)\] are \[{{E}_{x}}=6-8\times (1)=-2\] \[{{E}_{y}}=-8(1)-8+6(1)=-10\] \[{{E}_{z}}=6\times 1=6\] So, \[{{E}_{net}}=\sqrt{{{(-2)}^{2}}+{{(10)}^{2}}+{{(6)}^{2}}}\] \[=\sqrt{4+100+36}=\sqrt{40}\Rightarrow \sqrt{35\times 4}\] \[=2\sqrt{35}N/C\] So, \[F=q\,{{E}_{net}}=2(2\sqrt{35})=4\sqrt{35}N\]

"> ALIGN} & {{E}_{x}}\frac{\delta V}{\delta x}=6-8y \\ & {{E}_{y}}=\frac{\delta V}{\delta y}=-8x-8+6z \\ \end{align} \RIGHT\}\] (ii) \[{{E}_{z}}=6y\] Above values of \[{{E}_{x}},{{E}_{y}}\], and \[{{E}_{z}}\] at \[(1,\text{ }1,\text{ }1)\] are \[{{E}_{x}}=6-8\times (1)=-2\] \[{{E}_{y}}=-8(1)-8+6(1)=-10\] \[{{E}_{z}}=6\times 1=6\] So, \[{{E}_{net}}=\sqrt{{{(-2)}^{2}}+{{(10)}^{2}}+{{(6)}^{2}}}\] \[=\sqrt{4+100+36}=\sqrt{40}\Rightarrow \sqrt{35\times 4}\] \[=2\sqrt{35}N/C\] So, \[F=q\,{{E}_{net}}=2(2\sqrt{35})=4\sqrt{35}N\]

">
Mobile Technologies Web Technologies Digital Marketing Creative Design Career Talk Medical General Tech Interviews Internet Of Things

Embark on a journey of knowledge! Take the quiz and earn valuable credits.

Take A Quiz

Challenge yourself and boost your learning! Start the quiz now to earn credits.

Take A Quiz

Unlock your potential! Begin the quiz, answer questions, and accumulate credits along the way.

Take A Quiz

Popular Categories

Interview Questions
QuestionBank
Quantitative Aptitude
Engineering
Digital Marketing
Management
Govt Exams
Interviews
All

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 \[(1,\text{ }1,\text{ }1)\] is [NEET 2014]

NEET Physics in NEET . 3 months ago

  7   0   0   0   0 tuteeHUB earn credit +10 pts
5 Star Rating 1 Rating

[d] We know \[F=qE\] (i) \[E=-\frac{dv}{dr}\] \[\left. \begin{ALIGN} & {{E}_{x}}\frac{\delta V}{\delta x}=6-8y \\ & {{E}_{y}}=\frac{\delta V}{\delta y}=-8x-8+6z \\ \end{align} \RIGHT\}\] (ii) \[{{E}_{z}}=6y\] Above values of \[{{E}_{x}},{{E}_{y}}\], and \[{{E}_{z}}\] at \[(1,\text{ }1,\text{ }1)\] are \[{{E}_{x}}=6-8\times (1)=-2\] \[{{E}_{y}}=-8(1)-8+6(1)=-10\] \[{{E}_{z}}=6\times 1=6\] So, \[{{E}_{net}}=\sqrt{{{(-2)}^{2}}+{{(10)}^{2}}+{{(6)}^{2}}}\] \[=\sqrt{4+100+36}=\sqrt{40}\Rightarrow \sqrt{35\times 4}\] \[=2\sqrt{35}N/C\] So, \[F=q\,{{E}_{net}}=2(2\sqrt{35})=4\sqrt{35}N\]

Previous Next

Posted on 08 Aug 2024, this text provides information on NEET related to Physics in NEET. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

Take Quiz To Earn Credits!

Turn Your Knowledge into Earnings.

tuteehub_quiz

Tuteehub forum answer Answers

Post Answer

No matter what stage you're at in your education or career, TuteeHub will help you reach the next level that you're aiming for. Simply,Choose a subject/topic and get started in self-paced practice sessions to improve your knowledge and scores.

Explore Other Libraries

Blogs Tutorials Question Bank Feeds Careernews