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  • Solve this problem Equal charges Q are placed at the vertices B and C of an equilateral triangle ABC of side a. The magnitude of electric field at the point A is

Equal charges Q are placed at the vertices B and  C  of an equilateral triangle ABC of side a. The magnitude of electric field at the point A is

  • Option 1)

    \frac{Q}{4\pi \varepsilon _{0}a^{2}}

     

     

     

  • Option 2)

    \frac{\sqrt{2}Q}{4\pi \varepsilon _{0}a^{2}}

  • Option 3)

    \frac{\sqrt{3}Q}{4\pi \varepsilon _{0}a^{2}}

  • Option 4)

    \frac{Q}{2\pi \varepsilon _{0}a^{2}}

 

Answers (1)

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}}

 

 

As shown in figure Net electric field at A

E=\sqrt{E_{B}^{2}+E_{C}^{2}+2E_{B}E_{C}\cos 60}

E_{B}=E_{C}=\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q}{a^{2}}

So, E=\frac{\sqrt{3}Q}{4\pi \varepsilon _{0}a^{2}}


Option 1)

\frac{Q}{4\pi \varepsilon _{0}a^{2}}

 

 

 

Option 2)

\frac{\sqrt{2}Q}{4\pi \varepsilon _{0}a^{2}}

Option 3)

\frac{\sqrt{3}Q}{4\pi \varepsilon _{0}a^{2}}

Option 4)

\frac{Q}{2\pi \varepsilon _{0}a^{2}}

Posted by

Avinash

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