Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place
on the same straight line, and adjacent charges are equally separated by a distance d. Consider the net electrostatic force acting on each charge. Calculate the ratio of the largest to the smallest net force magnitude.
1 answer:
Answer:
Ans= 9
See attached picture for clearer solution.
Explanation:
The net electrostatic force acting on charge A = 2/ 2 + 2 /(2) 2 − 2 /(3) 2 = 2 / 2 (1 + 1/4 – 1/9 ) = 41/36 2/2 .
The net electrostatic force acting on charge B = 2/2 + 2/(2)2 − 2/2 = 1/4 2/d2 .
The net electrostatic force acting on charge C = 2/2 + 2/(2)2 + 2/2 = 2/2 (1 + 1 4 + 1) = 9/4 2/2 .
The net electrostatic force acting on charge D = 2/2+ 2 /(2)2 + 2/(3)2 = 2 /2 (1 + 1/4 + 1/9 ) = 49/36 2/ 2 .
The ratio of the largest to the smallest net force = 9/4*2/2 / 1/4 2/2 . = 9
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b) T_{o2} =1.58*10^5 K
Explanation:
a)= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h
b)
M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol
gas constant R = 8.31 j/mol.K
So,
multiply each side by M_{o2}, so we have
solving for temperature T_{o2}
In the question given,
T_{o2} =1.58*10^5 K