
Two identical copper spheres are separated by 1 m in vacuum. How many electrons would have to be removed from one sphere and added to another 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. \[6.25\times {{10}^{13}}\]
D. \[0.65\times {{10}^{13}}\]
Answer
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Hint: In this question we are asked to calculate the number of electrons to be removed from one of the two identical spheres so that they attract each other with a force of 0.9 N. We will be using Coulomb’s Law to calculate the desired result. Coulomb’s law states the relation between the force and two charged particles separated by some distance.
Formula Used:
\[F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
Where,
F is the force between the particles
k is the Coulomb's constant
q is the charge on the particle
r is the distance between two particles
Complete answer :
We know that the relation between force and two charged particles separated by a certain distance is given by Coulomb’s law. It states that, Force between the two particles is directly proportional to the product of charge of two particles and inversely proportional to the square of distance between them.
It is given as,
\[F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
Let us assume that charge on one particle \[{{q}_{1}}=x\] and \[{{q}_{2}}=-x\]. Also, we know that Coulomb’s constant k = \[9\times {{10}^{9}}\] \[N{{m}^{2}}/{{C}^{2}}\]
After substituting values
We get,
\[0.9=\dfrac{-9\times {{10}^{9}}\times {{x}^{2}}}{{{1}^{2}}}\]
Therefore,
\[{{x}^{2}}=\dfrac{1}{{{10}^{5}}}\]
It can be written as
\[{{x}^{2}}={{10}^{-5}}C\]
Now we know that 1 Coulomb has \[6.25\times {{10}^{18}}\] electrons.
Therefore,
\[6.25\times {{10}^{18}}\times {{10}^{-5}}=6.25\times {{10}^{13}}\]
Therefore, \[{{10}^{-5}}\] Coulombs will have \[6.25\times {{10}^{13}}\]. Therefore, \[6.25\times {{10}^{13}}\] electrons should be removed from one sphere and added to another to make the attractive force F = 0.9N.
Therefore, the correct answer is option C.
Note:
Coulomb’s Law states that the electrical force between two bodies is directly proportional to the product of the quantity of the charge on bodies and inversely proportional to the distance of separation between them. Coulomb’s law can provide a relatively simple derivation of Gauss’s law for general cases. The vector form of Coulomb’s law specifies the direction of electric field due to charges.
Formula Used:
\[F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
Where,
F is the force between the particles
k is the Coulomb's constant
q is the charge on the particle
r is the distance between two particles
Complete answer :
We know that the relation between force and two charged particles separated by a certain distance is given by Coulomb’s law. It states that, Force between the two particles is directly proportional to the product of charge of two particles and inversely proportional to the square of distance between them.
It is given as,
\[F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
Let us assume that charge on one particle \[{{q}_{1}}=x\] and \[{{q}_{2}}=-x\]. Also, we know that Coulomb’s constant k = \[9\times {{10}^{9}}\] \[N{{m}^{2}}/{{C}^{2}}\]
After substituting values
We get,
\[0.9=\dfrac{-9\times {{10}^{9}}\times {{x}^{2}}}{{{1}^{2}}}\]
Therefore,
\[{{x}^{2}}=\dfrac{1}{{{10}^{5}}}\]
It can be written as
\[{{x}^{2}}={{10}^{-5}}C\]
Now we know that 1 Coulomb has \[6.25\times {{10}^{18}}\] electrons.
Therefore,
\[6.25\times {{10}^{18}}\times {{10}^{-5}}=6.25\times {{10}^{13}}\]
Therefore, \[{{10}^{-5}}\] Coulombs will have \[6.25\times {{10}^{13}}\]. Therefore, \[6.25\times {{10}^{13}}\] electrons should be removed from one sphere and added to another to make the attractive force F = 0.9N.
Therefore, the correct answer is option C.
Note:
Coulomb’s Law states that the electrical force between two bodies is directly proportional to the product of the quantity of the charge on bodies and inversely proportional to the distance of separation between them. Coulomb’s law can provide a relatively simple derivation of Gauss’s law for general cases. The vector form of Coulomb’s law specifies the direction of electric field due to charges.
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