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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.

Respuesta :

syed514
 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