Question

Two point charge $ \;{q_1} = 3 \times {10^{ - 4}}\,C $ and $ {q_2} = 5 \times {10^{ - 4}}C $ are located at $ \left( {3,5,1} \right) $ and $ \left( {1,3,2} \right)m $ . Find $ \overrightarrow {{F_{12}}} $ and $ \overrightarrow {{F_{21}}} $ using the vector form of Coulomb's law. Also, find their magnitude.

Answer 1

Hint : The force between two charges is calculated using coulomb’s law which is proportional to the product of the two charges and inversely proportional to the square of the distance between them. The vector between two points can be found using the difference of the coordinates of the two points

Formula used: Coulomb's law: $ F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}} $ where $ F $ is the force acting between two charged particles of charge $ {q_1} $ and $ {q_2} $ which have a distance $ r $ between them.

Complete step by step answer:
We’ve been given that two point charges $ \;{q_1} = 3 \times {10^{ - 4}}\,C $ and $ {q_2} = 5 \times {10^{ - 4}}C $ are located at $ \left( {3,5,1} \right) $ and $ \left( {1,3,2} \right)m $ .
The vector form of Coulomb’s law can be written as:
 $ F = \dfrac{{k{q_1}{q_2}}}{{|r{|^2}}}\hat r $
Where $ \hat r $ is the unit vector joining in the two charges and $ |r| $ is the magnitude of the vector.
For two-point charges located at $ \left( {3,5,1} \right) $ and $ \left( {1,3,2} \right)m $ , the vector indirection of the second point charge from the first point charge will be
 $ {\vec r_{12}} = \left( {(1 - 3)\hat i + (3 - 5)\hat j + (2 - 1)\hat k} \right) $
 $ \Rightarrow {\vec r_{12}} = - 2\hat i - 2\hat j + \hat k $
The magnitude of this vector will be
 $ \left| {{{\vec r}_{12}}} \right| = \sqrt { - {2^2} + {{( - 2)}^2} + {{(1)}^2}} $
 $ \Rightarrow \left| {{{\vec r}_{12}}} \right| = 3\,m $
Hence the unit vector $ \hat r $ will be
 $ \hat r = \dfrac{{\vec r}}{{|r|}} $
 $ \Rightarrow \hat r = \dfrac{{ - 2\hat i - 2\hat j + \hat k}}{3} $
Substituting $ \;{q_1} = 3 \times {10^{ - 4}}\,C $ and $ {q_2} = 5 \times {10^{ - 4}}C $ and $ |r| = 3 $ and $ \hat r = \dfrac{{ - 2\hat i - 2\hat j + \hat k}}{3} $ in $ F = \dfrac{{k{q_1}{q_2}}}{{|r{|^2}}}\hat r $ , we get
 $ \overrightarrow {{F_{12}}} = \dfrac{{9 \times {{10}^9} \times \;3 \times {{10}^{ - 4}}\, \times 5 \times {{10}^{ - 4}}C}}{{{3^2}}}\left( { - 2\hat i - 2\hat j + \hat k} \right) $
Which on simplifying give us
 $ \overrightarrow {{F_{12}}} = 50\left( { - 2\hat i - 2\hat j + \hat k} \right) $
The force $ \overrightarrow {{F_{21}}} $ will be the additive inverse of $ \overrightarrow {{F_{12}}} $ so
 $ \overrightarrow {{F_{21}}} = - \overrightarrow {{F_{12}}} $
Which can be written as
 $ \overrightarrow {{F_{21}}} = - 50\left( { - 2\hat i - 2\hat j + \hat k} \right) $ or,
 $ \overrightarrow {{F_{21}}} = 50\left( {2\hat i + 2\hat j - \hat k} \right) $
Both the force vectors will have the same magnitude which we can calculate as
 $ |{F_{12}}| = |{F_{21}}| = 50 \times \sqrt {{2^2} + {2^2} + {1^2}} $
Hence the magnitude of the force between the two charges will be
 $ |{F_{12}}| = |{F_{21}}| = 150\,N $ .

Note:
The two forces $ \overrightarrow {{F_{12}}} $ and $ \overrightarrow {{F_{21}}} $ will have the same magnitude as they are acting on each other and according to Newton’s third law, action and reaction forces have the same magnitude. However, the direction of forces will be opposite to each other.