The diagram below shows electric field lines around two isolated point changes $\mathrm{P}$ and $\mathrm{Q}$. At $X$ the field strength is zero. Which of the following is true ?

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Aicharya · Accepted Answer

From the diagram, we conclude that (1) $\mathrm{Q}$ is larger charge because $\mathrm{X}$ is closer to $\mathrm{P}$ than Q. When $E_{P}=E_{Q}=\frac{q_{P}}{4 \pi \varepsilon_{0} r_{P}^{2}}=\frac{q_{Q}}{4 \pi \varepsilon_{0} r_{Q}^{2}} \therefore \frac{q_{P}}{q_{Q}}=\left(\frac{r_{P}}{r_{Q}}\right)^{2}$ Since $r_{Q} > r_{p}$, then $q_{Q} > q_{p} .(2)$ The field strength is always proportional to the inverse squared distance from $X$ as it can be readily seen from the expression in $(1),(3) .$ The potential at $\mathrm{P}$ or $\mathrm{Q}$ is infinitely large since the potential at the location of an isolated charge is theoretically infinite. (4) Both charges are positive since the field lines are radially outward from the charges. Hence (d) is correct. It may additionally be noted ~

The diagram below shows electric field lines around two isolated point changes \(\mathrm{P}\) and…

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