Question 12

(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^{−7} C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer

(a) Charge on sphere A, q_{A} = Charge on sphere B,

q_{B} = 6.5 × 10^{−7} C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres,

Where, ∈0 = Free space permittivity

=9x10^{9}N m^{2 }C^{2}

∴

=1.52X10^{-2}N Therefore, the force between the two spheres is 1.52 × 10^{−2} N.

(b) As given in the question

After doubling the charge of sphere,

charge on sphere A would be,

q_{A} = 2 × 6.5 × 10^{−7} C

and charge on sphere B would be,

q_{B} = 1.3 × 10^{−6} C

The distance between the spheres is halved as given:

∴

Now,force of repulsion between the two spheres,

= 16 × 1.52 × 10^{−2}

= 0.243 N

Therefore, the force of repulsion between the two spheres is 0.243 N.

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Swati
2019-07-06 06:38:47

Thanks

Megha
2019-06-05 23:15:39

Thanx it's really helpful

Jashwant
2019-03-31 10:14:26

Thanks

Anisha
2018-08-19 18:12:26

Plsss tell me ki all NCERT k questions solve kaha or konsi web pr hai

Melvin
2018-04-10 11:45:37

Thank you so much

Swati
2017-12-11 10:35:43

thank u not only for this question but for all the questions..thank u so much

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