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(A) $1$. Electric field at $O$: The electric field due to pairs of opposite charges at $O$ are:- Due to $A(+2q)$ and $D(-2q)$: $E_{AD} = \frac{1}{4\pi

Vedclass · Accepted Answer

$(A, B, C)$

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(A) $1$. Electric field at $O$: The electric field due to pairs of opposite charges at $O$ are:- Due to $A(+2q)$ and $D(-2q)$: $E_{AD} = \frac{1}{4\pi\varepsilon_0} \frac{2q}{L^2} + \frac{1}{4\pi\varepsilon_0} \frac{2q}{L^2} = 4K$ (along $OD$)- Due to $F(+q)$ and $C(-q)$: $E_{FC} = \frac{1}{4\pi\varepsilon_0} \frac{q}{L^2} + \frac{1}{4\pi\varepsilon_0} \frac{q}{L^2} = 2K$ (along $OD$)- Due to $B(+q)$ and $E(-q)$: $E_{BE} = \frac{1}{4\pi\varepsilon_0} \frac{q}{L^2} + \frac{1}{4\pi\varepsilon_0} \frac{q}{L^2} = 2K$ (along $OD$)Total electric field $E_O = 4K + 2K = 6K$ along $OD$. Thus,$(A)$ is correct.$2$. Potential at $O$: $V_O = \sum \frac{kq_i}{r_i} = \frac{1}{4\pi\varepsilon_0 L} (2q - 2q + q - q + q - q) = 0$. Thus,$(B)$ is correct.$3$. Potential on line $PR$: The line $PR$ is the perpendicular bisector of the line joining the charges. For any point on $PR$,the distance to $+q$ and $-q$ charges is equal,making the net potential zero. Thus,$(C)$ is correct.$4$. Potential on line $ST$: The potential varies along $ST$ because it is not an equipotential line. Thus,$(D)$ is incorrect.Therefore,the correct statements are $(A, B, C)$.

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