Question

Two point charges $ + 4\mu C $ and $ + 2\mu C $ repel each other with a force $ 80N $ what is the distance between the charges?
(A) $ 3cm $
(B) $ 0.3cm $
(C) $ 3 \times {10^{ - 2}}cm $
(D) $ 3 \times {10^{ - 3}}cm $

Answer 1

Hint: We are asked to find the distance between the two given point charges. We can start by writing down the information given in the question. Then we can proceed to find the distance between the charges using the formula that gives us the force between two point charges given by Coulomb's law.
The formula for the force between two point charges is given by the coulomb's law and it is as follows
 $ F = \dfrac{{k{Q_1}{Q_2}}}{{{r^2}}} $
Where $ {Q_1} $ is the first point charge and $ {Q_2} $ is the second point charge
 $ r $ is the distance separating the two point charges
 $ k = \dfrac{1}{{4\pi {\varepsilon _0}}} $ .

Complete answer:
We can start to attempt this question by writing down the values or data given in the question
The value of the first point charge is given as $ {Q_1} = + 4\mu C $
The value of the second point charge is given as $ {Q_2} = + 2\mu C $
The repulsive force between them is given as $ F = 80N $
The value of the constant k is $ k = 9 \times {10^9}kg{m^3}{C^{ - 2}}{s^{ - 2}} $
Now let us remember the formula of coulomb's law, which is $ F = \dfrac{{k{Q_1}{Q_2}}}{{{r^2}}} $
We can substitute the values given and find the value of the distance between the point charges as
 $ r = \sqrt {\dfrac{{k{Q_1}{Q_2}}}{F}} = \sqrt {\dfrac{{9 \times {{10}^9} \times 4 \times {{10}^{ - 6}} \times 2 \times {{10}^{ - 6}}}}{{80}}} = \sqrt {9 \times {{10}^{ - 4}}} = 3 \times {10^{ - 2}}m $
In conclusion, the right answer is option (A) $ 3cm $ .

Note:
The Coulomb's law gives us the force between two point charges separated by a distance. Coulomb's law is an inverse square law. The statement of coulomb's law is the force of attraction of repulsion between two point charges separated by a distance is directly proportional to the product of the charges and inversely proportional to the square of the distance separating them”.