q = 25nC, L = 5cm. Calculate: a.The force in the load of the lower left corner,
ID: 1651450 • Letter: Q
Question
q = 25nC, L = 5cm. Calculate:
a.The force in the load of the lower left corner,
b.The electric field at the midpoint of the horizontal upper segment,
c.The potential energy of the load in the lower left corner,
d.The electrostatic energy of the quadrupole
.
l. -q -q q=25nC, L-5cm. Calcule: a. La fuerza en la carga de la esquina inferior izquierda, b. El campo eléctrico en el punto medio del segmento superior horizontal, c. La energía potencial de la carga de la esquina inferior izquierda, d. La energía electrostática del cuadrupolo.Explanation / Answer
A) The force in the load of lower left corner
= kq*-q/L^2 cos 45 degree + k*-q*-q/(2 L^2)
= 9e9*-25e-9^2 /0.05^2 cos 45 degree + 9e9*25e-9^2/(2*0.05^2)
= - 0.000466 N
Hence force is 0.000466 N answer
B) Electric field E = 2* kq/(L/2)^2 -2kq/(L^2 +0.25L^2) *0.5/sqrt(1.25)
= 2*9e9*25e-9/0.025^2 - 2*9e9*25e-9/(0.05^2 + 0.025^2)*0.5/sqrt(1.25)
= 655601 N/C answer
C) PE of lower left = - 2 kq^2 /L + kq^2 /(L sqrt 2)
= - 2*9e9*25e-9^2 /0.05 + 9e9*25e-9^2 /(1.41*0.05)
= - 0.000145 J
D) Energy of the system = summation kq1q2/r
= - 4*9e9*25e-9^2 /0.05 + 2*9e9*25e-9^2 /(1.41*0.05)
= - 0.000290 J
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