Answer
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Hint
Here, we need to find the value of forces between the two charges, we will use the coulomb’s law i.e. $F = \dfrac{{{q_1}{q_2}}}{{{r^2}}}$. We find the forces on 1st charge due to second charge and then find the force on second charge due to 1st charge. Then we find the ratio between the two charges.
Complete step by step answer
Here two charges are given +1microcoulomb and +5microcoulomb. To find the force of attraction or repulsion between them we use coulomb’s law. Which states that the force between the two charges is directly proportional to the product of two charges and inversely proportional to the square of distance between them i.e. $F = \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Now first we find the force on +1 microcoulomb charge due to +5 microcoulomb charge i.e. $F_{12}$
Using coulomb’s law, we get
$ \Rightarrow {F_{12}} = \dfrac{{\left( { + 1} \right)\left( { + 5} \right)}}{{{r^2}}}$, r is the distance between these two charges …………………………. (1)
Similarly, we can write the force on +5 microcoulomb charge due to +1 microcoulomb charge i.e. $F_{21}$
By using Coulomb’s law, we get
$ \Rightarrow {F_{21}} = \dfrac{{\left( { + 5} \right)\left( { + 1} \right)}}{{{r^2}}}$, here also r is the distance between these charges ………………………………… (2)
Now, in order to find the ratio between both the forces we can divide the equation (1) by equation (2), we get
$ \Rightarrow \dfrac{{{F_{12}}}}{{{F_{21}}}} = \dfrac{{\dfrac{{\left( { + 1} \right)\left( { + 5} \right)}}{{{r^2}}}}}{{\dfrac{{\left( { + 5} \right)\left( { + 1} \right)}}{{{r^2}}}}} = \dfrac{1}{1}$
Hence, the ratio between the forces is 1:1.
Thus, option (A) is correct.
Note
Coulomb’s law states that the force acting between the two charges is directly proportional to the product of charges between them and inversely proportional to the square of distance between them i.e. $F = \dfrac{{{q_1}{q_2}}}{{{r^2}}}$, r is the distance between the two charges. If this force is negative it indicates that the force between the two charges is attractive and if force between the two charges is positive it indicates that the force is repulsive.
Here, we need to find the value of forces between the two charges, we will use the coulomb’s law i.e. $F = \dfrac{{{q_1}{q_2}}}{{{r^2}}}$. We find the forces on 1st charge due to second charge and then find the force on second charge due to 1st charge. Then we find the ratio between the two charges.
Complete step by step answer
Here two charges are given +1microcoulomb and +5microcoulomb. To find the force of attraction or repulsion between them we use coulomb’s law. Which states that the force between the two charges is directly proportional to the product of two charges and inversely proportional to the square of distance between them i.e. $F = \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Now first we find the force on +1 microcoulomb charge due to +5 microcoulomb charge i.e. $F_{12}$
Using coulomb’s law, we get
$ \Rightarrow {F_{12}} = \dfrac{{\left( { + 1} \right)\left( { + 5} \right)}}{{{r^2}}}$, r is the distance between these two charges …………………………. (1)
Similarly, we can write the force on +5 microcoulomb charge due to +1 microcoulomb charge i.e. $F_{21}$
By using Coulomb’s law, we get
$ \Rightarrow {F_{21}} = \dfrac{{\left( { + 5} \right)\left( { + 1} \right)}}{{{r^2}}}$, here also r is the distance between these charges ………………………………… (2)
Now, in order to find the ratio between both the forces we can divide the equation (1) by equation (2), we get
$ \Rightarrow \dfrac{{{F_{12}}}}{{{F_{21}}}} = \dfrac{{\dfrac{{\left( { + 1} \right)\left( { + 5} \right)}}{{{r^2}}}}}{{\dfrac{{\left( { + 5} \right)\left( { + 1} \right)}}{{{r^2}}}}} = \dfrac{1}{1}$
Hence, the ratio between the forces is 1:1.
Thus, option (A) is correct.
Note
Coulomb’s law states that the force acting between the two charges is directly proportional to the product of charges between them and inversely proportional to the square of distance between them i.e. $F = \dfrac{{{q_1}{q_2}}}{{{r^2}}}$, r is the distance between the two charges. If this force is negative it indicates that the force between the two charges is attractive and if force between the two charges is positive it indicates that the force is repulsive.
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