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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, qA = Charge on sphere B,

qB = 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

fraction numerator 1 over denominator 4 straight pi element of subscript 0 end fraction=9x109N m2 C2

=1.52X10-2N 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,

qA = 2 × 6.5 × 10−7 C

and charge on sphere B would be,

qB = 1.3 × 10−6 C

The distance between the spheres is halved as given:

r space equals fraction numerator 0.5 over denominator 2 end fraction equals 0.25 space m

Now,force of repulsion between the two spheres,

F equals fraction numerator q subscript A q subscript B over denominator 4 straight pi element of subscript 0 straight r squared end fraction

equals fraction numerator 9 x 10 to the power of 9 x 1.3 x 10 to the power of minus 6 end exponent x 1.3 x 10 to the power of minus 6 end exponent over denominator left parenthesis 0.25 right parenthesis squared end fraction

= 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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