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  • Please,please help me As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is

As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is

  • Option 1)

    \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{2q}{\sqrt{a^{2}+b^{2}}}

     

     

     

  • Option 2)

    \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{q}{\sqrt{a^{2}+b^{2}}}

  • Option 3)

    \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{(-q)}{\sqrt{a^{2}+b^{2}}}

  • Option 4)

    Zero

 

Answers (1)

best_answer

As we learned

 

Negative Potential -

Due to Negative charge.

-

 

 Potential at A = Potential due to (+q) charge + Potential due to (– q) charge

=\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{q}{\sqrt{a^{2}+b^{2}}}+\frac{1}{4\pi \varepsilon _{0}}\frac{(-q)}{\sqrt{a^{2}+b^{2}}}=0


Option 1)

\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{2q}{\sqrt{a^{2}+b^{2}}}

 

 

 

Option 2)

\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{q}{\sqrt{a^{2}+b^{2}}}

Option 3)

\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{(-q)}{\sqrt{a^{2}+b^{2}}}

Option 4)

Zero

Posted by

Aadil

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