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Two point charges q_{1}\left ( \sqrt{10} \mu C\right ) and q_{2}\left ( -25 \mu C\right ) are placed on the x-axis at x = 1m and x = 4m respectively. The electric field (in V/m) at a point y = 3m on y - axis is,

\left [ take \frac{1}{4\pi\epsilon _{0}} = 9\times 10^{9}Nm^{2}C^{-2} \right ]

  • Option 1)

    \left ( -81\widehat{i}+81\widehat{j} \right )\times 10^{2}

  • Option 2)

    \left ( -81\widehat{i}-81\widehat{j} \right )\times 10^{2}

  • Option 3)

    \left ( -63\widehat{i}+27\widehat{j} \right )\times 10^{2}

  • Option 4)

    \left ( 63\widehat{i}-27\widehat{j} \right )\times 10^{2}

Answers (1)

best_answer

 

Superposition of Electric field -

The resultant electric field at any point is equal to the vector sum of all the electric fields.

 

- wherein

\vec{E}=\vec{E_{1}}+\vec{E_{2}}+\vec{E_{3}}+\cdots\vec{E_{n}}

\cos \theta _{1}=\frac{1}{\sqrt{10}}

\sin \theta _{1}=\frac{3}{\sqrt{10}}

\cos \theta _{2}=\frac{4}{5}

\sin \theta _{2}=\frac{3}{5}

\left | \vec{E_{1}} \right |=\frac{1}{4\pi\varepsilon _{0}} \frac{\sqrt10\times 10^{-6}}{10}=9\sqrt{10}\times 10^{2}

\left | \vec{E_{2}} \right |=\frac{1}{4\pi\varepsilon _{0}} \frac{25\times 10^{-6}}{25}=9\times 10^{3} \: \: V/m

\vec{E} = \vec{E_{1}}+\vec{E_{2}}

     =9\sqrt{10}\times 10^{2}[\cos \theta _{1}(-\hat{i})+\sin \theta _{1}\hat{j}] + 9\times 10^{3}[\cos \theta _{1}(\hat{i})-\sin \theta _{2}\hat{j}]

Substituting values of \cos \theta and  \sin \theta

\vec{E} = 63\hat{i}-27\hat{j}

 

 

 

 


Option 1)

\left ( -81\widehat{i}+81\widehat{j} \right )\times 10^{2}

Option 2)

\left ( -81\widehat{i}-81\widehat{j} \right )\times 10^{2}

Option 3)

\left ( -63\widehat{i}+27\widehat{j} \right )\times 10^{2}

Option 4)

\left ( 63\widehat{i}-27\widehat{j} \right )\times 10^{2}

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