Answer:
[tex]F_1=540*10^{-12} N[/tex]
[tex]F_2=-1080 * 10^{-12} N[/tex]
[tex]F_3=-2268 * 10^{-12} N[/tex]
Explanation:
From the question we are told that
Charge on objects
[tex]Q_1=4.0 * 10^6 C[/tex]
[tex]Q_2= 6.0 * 10^6 C[/tex]
[tex]Q_3= +9.0 * 10^6 C[/tex]
Distance apart [tex]d=0.50m[/tex]
Generally the equation for Force on individual Charge is mathematically given by
For Q_1
[tex]F_1=\frac{d*Q_1Q_2}{r^2}+\frac{ d*Q_1Q_3}{4r^2}[/tex]
[tex]F_1=\frac{k*(4.0 * 10^6 C)(6.0 * 10^6 C)}{0.5^2}+\frac{ d*(4.0 * 10^6 C)( +9.0 * 10^6 C)}{4(0.5)^2}[/tex]
[tex]F_1=540*10^{-12}[/tex]
For Q_2
[tex]F_2= \frac{k}{0.5^2[Q_1 + Q_3]Q_2}[/tex]
[tex]F_2= \frac{k}{0.5^2[4.0 * 10^6 C + 9.0 * 10^6 ]*6.0 * 10^6 C}[/tex]
[tex]F_2=-1080 * 10^{-12} N[/tex]
For Q_3
[tex]F_3=\frac{kQ_3}{r^2[Q_1/4+Q_2}[/tex]
[tex]F_3=\frac{k(+9.0 x 10^6 C)}{(0.5)^2[(4.0 x 10^6 )/4+(6.0 x 10^6 C)}[/tex]
[tex]F_3=-2268 * 10^{-12} N[/tex]
