Question

Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7C6.5×10−7C? The radii of A and B are negligible compared to the distance of separation.​

Answer 1

Answer:

Step-by-step explanation:

                            use  F = kq/r² formula directly u will get it

                                 k= 9 x 10 to the power 9

     hope it helps u mate.................

Answer 2

(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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