Two point charges q_1 = 2.10 nc and q_2 = -6.20 nc are 0.100 m apart. point a is midway between them; point b is 0.080 m from q_1 and 0.060 m from q_2 (the figure (figure 1) ). take the electric potential to be zero at infinity.

 1)- Va= V(q1 on A)+V(q2 on A)= q1*9*10^9/(d1) +q2*9*10^9/(d2)if these particles in the air = 2.4*10^-9 * 9*10^9 /0.05 + -6.5*10^-9 * 9*10^9 / 0.05 = -738 volt. _ where : d1= the distance between q1 and A. d2 = the distance between q2 and A. 
2)- Vb= V(q1 on B)+V(q2 on A)= q1*9*10^9/(d3)+ q2*9*10^9/(d4) = 2.4*10^-9 * 9*10^9/0.08 + -6.5*10^-9 * 9*10^9/0.06 = -705volt 3)- work = qc*(Va-Vb)= 2.8*10^-9 *(-738-(-705))= -92.4*10^-9 J

Earthplanet yplanet y5=earth     *many different equationsplanet x1.5=earthplanet x is two times planet yobject 5,2500 mass5,1,3,2,4object 1 is four times obect 4