Electric Charges and Fields - Result Question 10

10. When air is replaced by a dielectric medium of force constant $K$, the maximum force of attraction between two charges, separated by a distance

(a) decreases K-times

[1999]

(b) increases K-times

(c) remains unchanged

(d) becomes $\frac{1}{K^{2}}$ times

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Answer:

Correct Answer: 10. (a)

Solution:

  1. (a) In air, $F _{\text{air }}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^{2}}$

In medium, $F_m=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{K r^{2}}$

$\therefore \frac{F_m}{F _{\text{air }}}=\frac{1}{K} \Rightarrow F_m=\frac{F _{\text{air }}}{K}$ (decreases K-times)