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  • Solve it, ABC is an equilateral triangle. Charges +qare placed at each corner. The electric intensity at Owill be

ABC is an equilateral triangle. Charges +q are placed at each corner. The electric intensity at O will be

  • Option 1)

    \frac{1}{4\pi \varepsilon _{0}}\frac{q}{r^{2}}

     

     

  • Option 2)

    \frac{1}{4\pi \varepsilon _{0}}\frac{q}{r}

  • Option 3)

    Zero

  • Option 4)

    \frac{1}{4\pi \varepsilon _{0}}\frac{3q}{r^{2}}

 

Answers (2)

best_answer

As we learned

 

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}}

 

 


Option 1)

\frac{1}{4\pi \varepsilon _{0}}\frac{q}{r^{2}}

 

 

Option 2)

\frac{1}{4\pi \varepsilon _{0}}\frac{q}{r}

Option 3)

Zero

Option 4)

\frac{1}{4\pi \varepsilon _{0}}\frac{3q}{r^{2}}

Posted by

Avinash

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Let k = 1/4π€°

The electric field intensity (E) at O due to A is |E1 |= k (q/r^2) [directed towards midpoint of BC]——-(1)

And electric field intensity (E) at O due to B is |E2| = k (q/r^2) [directed towards midpoint of AC] ———(2)

The electric field intensity (E) at O due to C is |E3| = k (q/r^2) [directed towards midpoint of AB] ——-(3)

From (1),(2),(3) we get the net electric intensity at O due to A,B,C combined is given by ,

E = E1 + E2 + E3

Note that electric field intensity E1,E2,E3 are vector quantities and they form a triad with angle between them = 120°, and since they are equal in magnitude they cancel out each other

Thus ,net electric intensity

E = 0

Posted by

Khushboo Kumbhare

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