JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 9)
If two charges q$$_1$$ and q$$_2$$ are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?
$$d\sqrt k$$
$$1\,.\,5d\sqrt k$$
$$k\sqrt d$$
$$2d\sqrt k$$
Explanation
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$$ \begin{aligned} & \text { dielectric constant }=\mathrm{K} \\\\ & F_{\text {medium }}=\frac{1}{4 \pi\left(\mathrm{K} \varepsilon_{0}\right)} \times \frac{q_{1} q_{2}}{d^{2}} \\\\ & \because \quad F_{\text {air }}=F_{\text {medium }} \\\\ & \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{\left(d_{\mathrm{air}}\right)^{2}}=\frac{1}{4 \pi\left(\mathrm{K} \varepsilon_{0}\right)} \frac{q_{1} q_{2}}{d^{2}} \\\\ & \therefore \quad d_{\text {air }}=d\sqrt{\mathrm{K}} \end{aligned} $$
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