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Three charged particles A, B, and C with charges -4q, 2q\: \: and\: -2q are present on the circumference of a circle of radius d. The charged particles A, C and the centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along the x-direction is: 
Option: 1 \frac{\sqrt{3}q}{4\pi\epsilon _0d^2}
 
Option: 2 \frac{3\sqrt{3}q}{4\pi\epsilon _0d^2}

Option: 3 \frac{\sqrt{3}q}{\pi\epsilon _0d^2}  

Option: 4 \frac{2\sqrt{3}q}{\pi\epsilon _0d^2}
 

Answers (1)

best_answer

 

 

  

 

Let E1 be the resultant electric field due to charge 2q and - 2q. The resultant E1 is in the direction of -2q

E_1=\frac{k2q}{d^2}+\frac{k2q}{d^2}=\frac{k4q}{d^2}

 E2 due to -4q is in the direction of -4q and magnitude is given by

E_2=\frac{k4q}{d^2}

The net electric field in the x-direction is 

\\E_x=E_1 cos30+E_2cos30=\frac{\sqrt{3}}{2}\times2\times\frac{4q}{4\pi\epsilon_0d^2}\\E_x=\sqrt{3}\frac{q}{\pi\epsilon_0d^2}

So the correct option is 3.

Posted by

vishal kumar

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