# Two charges of  -7 C  and  5 C are positioned on a line at points  5  and  -6 , respectively. What is the net force on a charge of  -2 C at  0 ?

Feb 17, 2018

F=7.54$\cdot {10}^{9} N$towards charge of 5C
coulomb's law F=$k . \frac{q Q}{r} ^ 2$ where k=$\frac{1}{4 \pi \epsilon}$=$9 \cdot {10}^{9}$
there are two forces involved one due to negative charge and one due to positive charge but both are in same direction i.e. towards positive charge because opposite charges attract while like charges repel .$\therefore$ net force is sum of both forces by principal of superimposition.
F=k[$2 \cdot \frac{5}{6} ^ 2 + 2 \cdot \frac{7}{5} ^ 2$]N=k{$\frac{10}{36} + \frac{14}{25}$]N=k[$\frac{754}{36 \cdot 25}$]N=$9 \cdot {10}^{9} \cdot \frac{754}{36 \cdot 25} N = 754 \cdot {10}^{9} / 100 N = 7.54 \cdot {10}^{9} N$
since, both forces were towards charge of 5C i.e. in -x direction. $\therefore$ net force will also be in the same direction.