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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Two point charges placed at a distance $r$ in air exert a force $F$ on each other. The distance at which they experience force $4 F$ when placed in a medium of dielectric constant $K = 16$ is -

Options:

Answer:

Solution:

Force in air at a distance $r,$

$F = \frac{q_{1} q_{2}}{4 π ϵ_{0} r^{2}}$ ......(i)

Let $r ’$ be the distance in the medium, therefore

$4 F = \frac{q_{1} q_{2}}{4 π ϵ_{0} (16) (r ’)^{2}}$ ......(ii)

From (i) $\div$ (ii)

$\frac{1}{4} = \frac{16 (r ’)^{2}}{r^{2}}$

$∴ r ’ = \frac{r}{8}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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