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(D) The electric field at the centre $O$ due to a charge $q'$ at distance $d$ is $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q'}{d^2}$.Let the angles o

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$\frac{\sqrt{3} q}{\pi \varepsilon_{0} d^{2}}$

Vedclass · Answer

(D) The electric field at the centre $O$ due to a charge $q'$ at distance $d$ is $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q'}{d^2}$.Let the angles of $A, B, C$ with the positive $x$-axis be $	heta_A = 30^{\circ}$,$	heta_B = 150^{\circ}$,and $	heta_C = -30^{\circ}$.The electric field components along the $x$-axis are:$E_{Ax} = \frac{1}{4 \pi \varepsilon_{0}} \frac{|-4q|}{d^2} \cos(180^{\circ} + 30^{\circ}) = \frac{4q}{4 \pi \varepsilon_{0} d^2} (-\cos 30^{\circ}) = -\frac{4q \sqrt{3}}{8 \pi \varepsilon_{0} d^2} = -\frac{\sqrt{3} q}{2 \pi \varepsilon_{0} d^2}$.$E_{Bx} = \frac{1}{4 \pi \varepsilon_{0}} \frac{2q}{d^2} \cos(150^{\circ}) = \frac{2q}{4 \pi \varepsilon_{0} d^2} (-\frac{\sqrt{3}}{2}) = -\frac{\sqrt{3} q}{4 \pi \varepsilon_{0} d^2}$.$E_{Cx} = \frac{1}{4 \pi \varepsilon_{0}} \frac{|-2q|}{d^2} \cos(180^{\circ} - 30^{\circ}) = \frac{2q}{4 \pi \varepsilon_{0} d^2} (-\cos 30^{\circ}) = -\frac{\sqrt{3} q}{4 \pi \varepsilon_{0} d^2}$.The total electric field along the $x$-direction is $E_x = E_{Ax} + E_{Bx} + E_{Cx} = -\frac{\sqrt{3} q}{2 \pi \varepsilon_{0} d^2} - \frac{\sqrt{3} q}{4 \pi \varepsilon_{0} d^2} - \frac{\sqrt{3} q}{4 \pi \varepsilon_{0} d^2} = -\frac{\sqrt{3} q}{\pi \varepsilon_{0} d^2}$.The magnitude is $\frac{\sqrt{3} q}{\pi \varepsilon_{0} d^2}$.

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 form an equilateral triangle as shown in the figure. The electric field at $O$ along the $x$-direction is

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