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Physics · Electrostatics

JEE Main 2024 — 31 January, Shift 2 — Question 32

Force between two point charges q1q_{1} and q2q_{2} placed in vacuum at ' rr ' cm apart is F. Force between them when placed in a medium having dielectric K=5\mathrm{K}=5 at ' r/5′cm\mathrm{r} / 5^{\prime} \mathrm{cm} apart will be:

  1. Option A:

    F/25F / 25

  2. Option B:

    5 F

    Correct
  3. Option C:

    F/5F / 5

  4. Option D:

    25 F

Answer: B

Step-by-step solution

In air F=14π∈0q1q2r2F=\frac{1}{4 \pi \in_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{2}}

In medium F′=14π( K∈0)q1q2(r′)2=254π(5ϵ0)q1q2(r)2=5 F\mathrm{F}^{\prime}=\frac{1}{4 \pi\left(\mathrm{~K} \in_{0}\right)} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\left(\mathrm{r}^{\prime}\right)^{2}}=\frac{25}{4 \pi\left(5 \epsilon_{0}\right)} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{(\mathrm{r})^{2}}=5 \mathrm{~F}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Physics
Chapter
Electrostatics
Topic
Properties of Electric Charge and Coulomb's Law

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