Two charges of # -2 C # and # 5 C# are positioned on a line at points # 4 # and # -1 #, respectively. What is the net force on a charge of # -9 C# at # 0 #?

1 Answer

See the answer below

Explanation:

The electrostatic force #F# between two point charges #Q_1# & #Q_2# separated by a distance #r# is given as

#F=\frac{KQ_1Q_2}{r^2}#

Where, #K# is constant & #K=9\times 10^9\ {Nm^2}/C^2#

Now, the electrostatic force #F_1# of repulsion between point charges #Q_1=-2C# & #Q_2=-9\ C# separated by a distance #r=4# is given as

#F_1=\frac{K(2)(9)}{4^2}=9/8 K#

Similarly, the electrostatic force #F_2# of attraction between point charges #Q_1=5C# & #Q_2=-9\ C# separated by a distance #r=1# is given as

#F_2=\frac{K(5)(9)}{1^2}=45K#

Since #F_1# & #F_2# on #-9C # are in the same direction hence the net force on #-9 C# considering direction of both the forces

#F_{\text{net}}=F_1+F_2#

#=9/8K+45K#

#=369/8 K#