Two particle of equal mass m and charge q are placed at a distance of 16 cm. They do not experience any force. The value of \( \frac{q}{m} \) is:
1. 1
2. \( \sqrt{\frac{\pi \varepsilon_0}{G}} \)
3. \( \sqrt{\frac{G}{4 \pi \varepsilon_0}} \)
4. \( \sqrt{4 \pi \varepsilon_0 G} \)
View Answer
For no net force, electrostatic force must equal gravitational force: \( \frac{q^2}{4 \pi \varepsilon_0 r^2} = \frac{G m^2}{r^2} \). Rearranging this formula yields \( \frac{q^2}{m^2} = 4 \pi \varepsilon_0 G \Rightarrow \frac{q}{m} = \sqrt{4 \pi \varepsilon_0 G} \).
Force between two identical spheres charged with same charge is \(F \). If 50% charge of one sphere is transferred to the other sphere then the new force will be:
1. \(\frac{3}{4}F\)
2. \(\frac{3}{8}F\)
3. \(\frac{3}{2}F\)
4. none of these
View Answer
Initially, \( F = k \frac{q^2}{r^2}\). When 50% of the charge of one sphere is transferred to the other, the charges become \(0.5q\) and
\(1.5q\). The new force is \(F' = k \frac{(0.5q)(1.5q)}{r^2} = 0.75 F = \frac{3}{4}F\).
Two point charges exert a force \(F_0\) on each other when placed in vacuum. Now the charges are increased to four times, separation between them is doubled and the system is placed is an insulating medium. Now they experience the same force. What should be the dielectric constant of the medium?
View Answer
Initial force in vacuum \(F_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . New force in medium \(F' = \frac{1}{4\pi\epsilon_0 K} \frac{(4q_1)(4q_2)}{(2r)^2} = \frac{16}{4K} \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . Since \(F' = F_0\), we have \(frac{4}{K} = 1\), so \(K = 4\).