First slide

Electric field

Question

Three charged particles A, B and C with charges $- 4 q, 2 q$ and $- 2 q$ are present on the circumference of a circle of radius $d$ . The charged particles A, C and centre O of the circle formed an equilateral triangle and point B is diametrically opposite to point C as shown in figure. Electric field at O along $X -$ direction is

NA

Solution

$E_{x} = \frac{K (4 q)}{d^{2}} \cos 30^{0} + \frac{K (2 q)}{d^{2}} \cos 30^{0} + \frac{K (2 q)}{d^{2}} \cos 30^{0} = \frac{K (4 q)}{d^{2}} \times (\frac{\sqrt{3}}{2}) + \frac{K (2 q)}{d^{2}} (\frac{\sqrt{3}}{2}) + \frac{K (2 q)}{d^{2}} (\frac{\sqrt{3}}{2}) = \frac{K .8 q}{d^{2}} (\frac{\sqrt{3}}{2}) = \frac{1}{4 π \in_{0}} \frac{8 q (\sqrt{3})}{d^{2} (2)} E_{x} = \frac{\sqrt{3} q}{π \in_{0} d^{2}}$

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