Two fixed charges 4,Q (positive) and Q (negat | vedclass

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Question

The resultant electric field will be zero at point closer to $\mathrm{B}$ and outside $\mathrm{AB}$ (by analysing directions of field and magnitudes)

Vedclass · Accepted Answer

$A$ and $B$ both

Vedclass · Answer

The resultant electric field will be zero at point closer to $\mathrm{B}$ and outside $\mathrm{AB}$ (by analysing directions of field and magnitudes)

If a positive charge is placed at $P$ and distributed, the positive charge either goes towards, $-Q$ or moves away from $-Q$ but won't oscillate

$\left(\because \frac{d^{2} v_{p}}{d x^{2}}>0\right)$

(unstable equilibrium) while negative charge oscillate

$\left(\because \frac{d^{2} U_{n}}{d^{2} x}\right.$$\left.=-\frac{\mathrm{d}^{2} \mathrm{U}_{\mathrm{P}}}{\mathrm{d}^{2} \mathrm{x}}<0\right)$ (Stable equilibrium)

Two fixed charges $4\,Q$ (positive) and $Q$ (negative) are located at $A$ and $B$, the distance $AB$ being $3$ $m$.

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