AP EAMCET · PHYSICS · Electrostatics
Two points charges are kept in air with a separation between them. The force between them is \(F_1\), if half of the space between the charges is filled with a dielectric constant 4 and the force between them is \(F_2\). If \(\frac{1}{3} \mathrm{rd}\) of the space between the charges is filled with dielectric of dielectric constant 9. Then \(\frac{F_1}{F_2}\) is
- A \(\frac{27}{64}\)
- B \(\frac{16}{81}\)
- C \(\frac{81}{64}\)
- D \(\frac{100}{81}\)
Answer & Solution
Correct Answer
(D) \(\frac{100}{81}\)
Step-by-step Solution
Detailed explanation
When dielectric of thickness \(t\) is introduced in two charges at distance \(r\), the effective force between the charges is given by \[ F=\frac{q_1 q_2}{4 \pi \varepsilon_0[r-t+t \sqrt{K}]^2} \] where, \(K=\) dielectric constant of medium In first case, \(t=r / 2\) and \(K=4\)…
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