# Two charges of  7 C  and  5 C are positioned on a line at points  2  and  -4 , respectively. What is the net force on a charge of  -4 C at  0 ?

Mar 6, 2016

$= 51.75 \cdot {10}^{9} N$

#### Explanation:

By Coulomb's law we Know that force between two charges Q and q situated at a distance r is given by the formula $F = k \frac{q Q}{r} ^ 2$,where k =Coulomb's constant=$9 \cdot {10}^{9} N {m}^{2} {C}^{-} 2$

Applying this formula the attractive force of 7C on -4C
=$9 \cdot {10}^{9} \cdot 4 \cdot \frac{7}{2} ^ 2 N = 63 \cdot {10}^{9} N$
The attractive force of 5C on -4C =$9 \cdot {10}^{9} \cdot 4 \cdot \frac{5}{4} ^ 2 N = 11.25 \cdot {10}^{9} N$

These two forces are oppositely directed so the net force will be towards right and its magnitude $= \left(63 - 11.25\right) \cdot {10}^{9} N = 51.75 \cdot {10}^{9} N$