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
Hint:
Here, we need to find the value of forces between the two charges, we will use the coulomb’s law i.e. [latex]F=\frac{q^1q^2}{r^2}[/latex]. 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 determine the force of attraction or repulsion between two charges, Coulomb’s law is applied. This law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance separating them.
i.e. [latex]F=\frac{q^1q^2}{r^2}[/latex]
Now first we find the force on +1 microcoulomb charge due to +5 microcoulomb charge i.e. [latex]F_{12}[/latex]
Using coulomb’s law, we get
⇒[latex]F12=\frac{(+1)(+5)}{r2}[/latex], 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. [latex]F_{21}[/latex]
By using Coulomb’s law, we get
⇒[latex]F_{21}=\frac{(+5)(+1)}{r^2}[/latex], here also r is the distance between these charges…………… (2)
To determine the ratio between the two forces, we can divide equation (1) by equation (2), yielding
⇒[latex]\frac{F_{12}}{F_{21}}=\frac{\frac{(+1)(+5)}{r^2}}{\frac{(+5)(+1)}{r^2}}=\frac{1}{1}[/latex]
Hence, the ratio between the forces is 1:1.
Thus, option (A) is correct.
Note:
Coulomb’s law describes the force between two charges, stating that it is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. i.e. [latex]F=\frac{q^1q^2}{r^2}[/latex], r is the distance between the two charges. A negative value for this force indicates an attractive interaction between the charges, while a positive value indicates a repulsive force between them.