Question

Two identical copper spheres are separated by Im in vacuum. How many electrons would have to
be removed from one sphere and added to the other so that they now attract each other with a
force of 0.9 N?
(A)
${6.25 \times 10}^{15 }$
(B)
${62.5 \times 10}^{15}$
(C) (D) 0.65 x 10​

Answer 1

$\mathtt{ \huge{ \fbox{Solution :)}}}$

Given ,

• The distance between two copper spheres = 1 m
• The electrostatic force between two copper spheres = 0.9

Let , the charges of two identical copper spheres be Q

We know that , the electrostatic force between two charges is given by

$\large \mathtt{ \fbox{F = k \frac{ Q_{1}Q_{2}}{ {(r)}^{2} } }}$

Where , K = coulombs constant which is 9 × (10)^9 Nm²/C²

Thus ,

$\sf \mapsto 0.9 = \frac{9 \times {(10)}^{9} \times {(Q)}^{2} }{ {(1)}^{2} } \\ \\ \sf \mapsto {(Q)}^{2} = \frac{0.9}{9 \times {(10)}^{9} } \\ \\ \sf \mapsto {(Q)}^{2} = \frac{1}{ {(10)}^{10} } \\ \\ \sf \mapsto Q = \sqrt{ {(10)}^{ -10} } \\ \\ \sf \mapsto Q = \sqrt{ { ({(10)}^{ - 5} )}^{2} } \\ \\ \sf \mapsto Q = {(10)}^{ - 5} \: \: coulomb$

We know that , the electric charge is an integral multiples of electrons

$\large \mathtt{ \fbox{Q = ne}}$

Thus ,

$\sf \mapsto {(10)}^{ - 5} = n \times 1.6 \times {(10)}^{ - 19} \\ \\ \sf \mapsto n = \frac{ {(10)}^{ - 5} }{1.6 \times {(10)}^{ - 19} } \\ \\ \sf \mapsto n = 0.625 \times {(10)}^{14} \\ \\ \sf \mapsto n = 6.25 \times {(10)}^{13}$

Hence , the number of electrons is 6.25 × (10)^13

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