Question

How many same number of excess electrons should be added to each neutral lead sphere to balance the force of gravity between them if their masses are 1kg each and they are separated by a distance of 1m

Answer 1

Answer:

Total number of excess electrons needed here to balance gravitational force is

$N = 5.38 \times 10^8$

Explanation:

As we know that gravitational force is balanced by electrostatic force between the two sphere

So we have

$\frac{Gm_1m_2}{r^2} = \frac{kq_1q_2}{r^2}$

here we know that

$m_1 = m_2 = 1 kg$

$q_1 = q_2 = q$

now we have

$G(1)(1) = k(q)(q)$

$6.67 \times 10^{-11} = (9\times 10^9)q^2$

$q = 8.608\times 10^{-11}$

now we have

$q = Ne$

$8.608 \times 10^{-11} = N(1.6 \times 10^{-19})$

$N = 5.38 \times 10^8$

#Learn

Topic : Electrostatic force

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