Three charged particle 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 centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is :

\sqrt 3 q \over \pi \varepsilon _0d^2

3\sqrt 3 q \over 4\pi \varepsilon _0d^2

\sqrt 3 q \over 4\pi \varepsilon _0d^2

2\sqrt 3 q \over \pi \varepsilon _0d^2

Electrostatics PYQ — JEE Main 2020 · 8 Jan · Shift 1 · Q47 with Solution

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Correct answer: C

Let the radius of the circle be $d$ , so each charge is at distance $d$ from the centre $O$ .
Since $A$ , $C$ and $O$ form an equilateral triangle, we have $\angle AOC = 60^\circ.$ Thus the radii $OA$ and $OC$ make angles $120^\circ$ and $60^\circ$ with the $+x$ -axis respectively, as suggested by the standard symmetric figure, while $B$ lies on the left end of the horizontal diameter.
Electric field magnitude at $O$ due to a charge $Q$ on the circumference is $E = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{d^2}.$ Direction:

for a negative charge, field at $O$ is towards the charge,
for a positive charge, field at $O$ is away from the charge.

We now find only the x-components.

Field due to charge $A = -4q$

Magnitude: $E_A = \frac{1}{4\pi\varepsilon_0}\frac{4q}{d^2}.$ Since $A$ is negative, field at $O$ is towards $A$ , i.e. along direction $120^\circ$ . Hence

= -\frac{2q}{4\pi\varepsilon_0 d^2}.$$ ### Field due to charge $C = -2q$ Magnitude: $$E_C = \frac{1}{4\pi\varepsilon_0}\frac{2q}{d^2}.$$ Since $C$ is negative, field at $O$ is towards $C$, i.e. along direction $60^\circ$. Thus $$E_{Cx} = E_C \cos 60^\circ = \frac{1}{4\pi\varepsilon_0}\frac{2q}{d^2}\left(\frac12\right) = \frac{q}{4\pi\varepsilon_0 d^2}.$$ ### Field due to charge $B = +2q$ Magnitude: $$E_B = \frac{1}{4\pi\varepsilon_0}\frac{2q}{d^2}.$$ Since $B$ is positive and lies on the negative x-axis, field at $O$ is away from $B$, i.e. along $+x$ direction. So $$E_{Bx} = \frac{2q}{4\pi\varepsilon_0 d^2}.$$ --- 4. Net electric field along x-direction: $$E_x = E_{Ax}+E_{Bx}+E_{Cx}.$$ Substitute: $$E_x = -\frac{2q}{4\pi\varepsilon_0 d^2}+\frac{2q}{4\pi\varepsilon_0 d^2}+\frac{q}{4\pi\varepsilon_0 d^2}$$ $$E_x = \frac{q}{4\pi\varepsilon_0 d^2}.$$ This does **not** match any option directly, so let us reconsider the geometry carefully. --- 5. In the usual figure for this problem, $A$ and $C$ are symmetric about the x-axis, and since $A,O,C$ form an equilateral triangle, each makes angle $30^\circ$ with the vertical diameter, so their x-components add using $\cos 30^\circ = \frac{\sqrt3}{2}$. Then: - For $A=-4q$, $$E_{Ax} = -\frac{1}{4\pi\varepsilon_0}\frac{4q}{d^2}\cos 60^\circ = -\frac{2q}{4\pi\varepsilon_0 d^2}$$ if interpreted one way, but from the figure-based orientation the effective horizontal contribution becomes $$E_{Ax}= -\frac{1}{4\pi\varepsilon_0}\frac{4q}{d^2}\cdot \frac{\sqrt3}{2} = -\frac{2\sqrt3 q}{4\pi\varepsilon_0 d^2}.$$ - For $C=-2q$, $$E_{Cx}= +\frac{1}{4\pi\varepsilon_0}\frac{2q}{d^2}\cdot \frac{\sqrt3}{2} = \frac{\sqrt3 q}{4\pi\varepsilon_0 d^2}.$$ - For $B=+2q$, from the figure its field is along $+x$: $$E_{Bx}=\frac{2q}{4\pi\varepsilon_0 d^2}.$$ However, even this mixed interpretation does not produce the listed answers unless the intended geometry gives only the resultant from the symmetric pair in x-direction: $$E_x = \frac{3\sqrt3 q}{4\pi\varepsilon_0 d^2}.$$ But the stored answer is option C: $$\boxed{\frac{\sqrt3 q}{\pi\varepsilon_0 d^2}}.$$ 6. Comparing with the options, option C equals $$\frac{\sqrt3 q}{\pi\varepsilon_0 d^2} = \frac{4\sqrt3 q}{4\pi\varepsilon_0 d^2}.$$ This would require the net x-component coefficient to be $4\sqrt3$, which is not obtained from the stated charge values and standard geometry. Therefore, based on electrostatics and the likely circle geometry, the stored answer appears inconsistent.

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View written solutionFree

Correct answer: C

Let the radius of the circle be $d$ , so each charge is at distance $d$ from the centre $O$ .
Since $A$ , $C$ and $O$ form an equilateral triangle, we have $\angle AOC = 60^\circ.$ Thus the radii $OA$ and $OC$ make angles $120^\circ$ and $60^\circ$ with the $+x$ -axis respectively, as suggested by the standard symmetric figure, while $B$ lies on the left end of the horizontal diameter.
Electric field magnitude at $O$ due to a charge $Q$ on the circumference is $E = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{d^2}.$ Direction:

for a negative charge, field at $O$ is towards the charge,
for a positive charge, field at $O$ is away from the charge.

We now find only the x-components.

Field due to charge $A = -4q$

Magnitude: $E_A = \frac{1}{4\pi\varepsilon_0}\frac{4q}{d^2}.$ Since $A$ is negative, field at $O$ is towards $A$ , i.e. along direction $120^\circ$ . Hence

= -\frac{2q}{4\pi\varepsilon_0 d^2}.$$ ### Field due to charge $C = -2q$ Magnitude: $$E_C = \frac{1}{4\pi\varepsilon_0}\frac{2q}{d^2}.$$ Since $C$ is negative, field at $O$ is towards $C$, i.e. along direction $60^\circ$. Thus $$E_{Cx} = E_C \cos 60^\circ = \frac{1}{4\pi\varepsilon_0}\frac{2q}{d^2}\left(\frac12\right) = \frac{q}{4\pi\varepsilon_0 d^2}.$$ ### Field due to charge $B = +2q$ Magnitude: $$E_B = \frac{1}{4\pi\varepsilon_0}\frac{2q}{d^2}.$$ Since $B$ is positive and lies on the negative x-axis, field at $O$ is away from $B$, i.e. along $+x$ direction. So $$E_{Bx} = \frac{2q}{4\pi\varepsilon_0 d^2}.$$ --- 4. Net electric field along x-direction: $$E_x = E_{Ax}+E_{Bx}+E_{Cx}.$$ Substitute: $$E_x = -\frac{2q}{4\pi\varepsilon_0 d^2}+\frac{2q}{4\pi\varepsilon_0 d^2}+\frac{q}{4\pi\varepsilon_0 d^2}$$ $$E_x = \frac{q}{4\pi\varepsilon_0 d^2}.$$ This does **not** match any option directly, so let us reconsider the geometry carefully. --- 5. In the usual figure for this problem, $A$ and $C$ are symmetric about the x-axis, and since $A,O,C$ form an equilateral triangle, each makes angle $30^\circ$ with the vertical diameter, so their x-components add using $\cos 30^\circ = \frac{\sqrt3}{2}$. Then: - For $A=-4q$, $$E_{Ax} = -\frac{1}{4\pi\varepsilon_0}\frac{4q}{d^2}\cos 60^\circ = -\frac{2q}{4\pi\varepsilon_0 d^2}$$ if interpreted one way, but from the figure-based orientation the effective horizontal contribution becomes $$E_{Ax}= -\frac{1}{4\pi\varepsilon_0}\frac{4q}{d^2}\cdot \frac{\sqrt3}{2} = -\frac{2\sqrt3 q}{4\pi\varepsilon_0 d^2}.$$ - For $C=-2q$, $$E_{Cx}= +\frac{1}{4\pi\varepsilon_0}\frac{2q}{d^2}\cdot \frac{\sqrt3}{2} = \frac{\sqrt3 q}{4\pi\varepsilon_0 d^2}.$$ - For $B=+2q$, from the figure its field is along $+x$: $$E_{Bx}=\frac{2q}{4\pi\varepsilon_0 d^2}.$$ However, even this mixed interpretation does not produce the listed answers unless the intended geometry gives only the resultant from the symmetric pair in x-direction: $$E_x = \frac{3\sqrt3 q}{4\pi\varepsilon_0 d^2}.$$ But the stored answer is option C: $$\boxed{\frac{\sqrt3 q}{\pi\varepsilon_0 d^2}}.$$ 6. Comparing with the options, option C equals $$\frac{\sqrt3 q}{\pi\varepsilon_0 d^2} = \frac{4\sqrt3 q}{4\pi\varepsilon_0 d^2}.$$ This would require the net x-component coefficient to be $4\sqrt3$, which is not obtained from the stated charge values and standard geometry. Therefore, based on electrostatics and the likely circle geometry, the stored answer appears inconsistent.

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