Elements with low specific heat values have which characteristics
1 answer:
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Answer:
5.634 N rightwards
Explanation:
qo = - 3 x 10^-7 C
q1 = - 9 x 10^-6 C
q2 = 10 x 10^-6 C
r1 = 7 cm = 0.07 m
r2 = 20 cm = 0.2 m
The force on test charge due to q1 is F1 which is acting towards right
According to the Coulomb's law
F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)
F1 = 4.959 N rightwards
The force on test charge due to q2 is F1 which is acting towards right
According to the Coulomb's law
F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)
F2 = 0.675 N rightwards
Net force on the test charge
F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards
