There are two charges +1 microcoulomb and +5microcoulomb. The ratio of the forces acting on them will be
(A) 1:1
(B) 1:25
(C) 1:5
(D) 5:1
Answer
542.1k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

