Numericals: 1. Two equal charges are kept at a distance 10 cm apart in air when they exert a force 49 dyne on each other. Find the magnitude of each charge.
2. The ratio of the amount of two charges is 2:3 and the distance between them is 5 cm. If the repulsive force acting between them is 96 dyne, then find the magnitude of each charge.
3. Two point charges, each having a charge 6.5×10-⁷ coulomb is separated by a distance of 75 cm in vacuum. Calculate the force of repulsion between them.
4. Relative permittivity of a medium is 3. Find the permittivity of the medium. 5. Amount of charge between two charged bodies and also the distance between them are made twice. What will be the change in force acting between them?
6. Two like charges ‘q1’ and ‘q2’ are kept at a distance ‘r’ apart and the force acting between them is ‘F’. Keeping the distance between them intact if the amount of ‘q1’ is made 4 times and the amount of ‘q2’ is made 5 times then what will be the amount of force acting between them?
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Wtrite in short okay plz
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1. F = 49 dyne
therefore F =49×10^-5N
R = 10 × 10^-2m.
F=k.q1.q2/R^2 . 49×10^-5 = 9×10^9×q2/10^-4
q=2.33×10^-6c aprox
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