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The excess number of electrons that must be placed on each of two small spheres spaced 3cm apart with force of repulsion b/w the spheres to be 10^-19 N is 1) 25 2) 225 3) 625 4) 1250 the correct answer is option 2. can you explain this answer?
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The excess number of electrons that must be placed on each of two smal...
Given that,
Distance between two spheres , r = 3 cm 
We are recharging into SI unit , i.e. in metre.
Now , r = 3 x 0.01 = 0.3 m
Force of repulsion , F = 10^-19N

We need to find the number of elections by using the formula q=ne.
So,first let us use coulombs formula to find the value of q(charge).
F = k q^2/r^2
10^-19 = 9 x 10^9 x q^2 / 0.09
q^2 = 10^-19 x 0.09 / 9 x 10^9
q^2 = 10^-28 x 10^-2 
q= 10^-15 C.
Substitute it in q=ne to get n,
n = q/e
n = 10^-15/1.6x10^-19
n = 625.

Therefore , option (3) is correct.
Community Answer
The excess number of electrons that must be placed on each of two smal...
Calculation of Excess Number of Electrons

Given:
- Distance between the two spheres = 3cm
- Force of repulsion between the spheres = 10^-19 N

To find: Excess number of electrons that must be placed on each sphere

Solution:

We can use Coulomb's law to find the excess number of electrons on each sphere. Coulomb's law states that the force of repulsion between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Mathematically,

F = k * (q1 * q2) / r^2

where,
- F = Force of repulsion
- k = Coulomb's constant (9 x 10^9 Nm^2/C^2)
- q1 and q2 = Charges on the spheres
- r = Distance between the spheres

We can rearrange the formula to find the charge on one sphere.

q1 = F * r^2 / (k * q2)

Substituting the given values,

q1 = 10^-19 * (0.03)^2 / (9 x 10^9 * q2)

q1 = 10^-28 / (3 x 10^-9 * q2)

q1 = 10^-19 / q2

We know that the excess charge on each sphere is the same. Let's assume that the excess charge on each sphere is n times the charge of an electron (e).

q1 = n * e

Substituting this in the above equation,

n * e = 10^-19 / q2

n = (10^-19 / q2) / e

We know that the charge of an electron is 1.6 x 10^-19 C. Substituting this value,

n = (10^-19 / q2) / (1.6 x 10^-19)

n = 0.625 / q2

We can rearrange this equation to find q2.

q2 = 0.625 / n

Substituting the given answer option 2,

n = 225

q2 = 0.625 / 225

q2 = 2.78 x 10^-3 C

The excess charge on each sphere is

q1 = n * e = 225 * 1.6 x 10^-19

q1 = 3.6 x 10^-17 C

Therefore, the excess number of electrons on each sphere is

n = q1 / e = 3.6 x 10^-17 / 1.6 x 10^-19

n = 225

Hence, the correct answer is option 2 (225).
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The excess number of electrons that must be placed on each of two small spheres spaced 3cm apart with force of repulsion b/w the spheres to be 10^-19 N is 1) 25 2) 225 3) 625 4) 1250 the correct answer is option 2. can you explain this answer?
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