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:
- (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)
