electricity
Publié le 19/12/2014
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Answer: As per the direction of the electric field, it is always from +ve to -ve charge.
So, for Q2,
the forces are from Q1 to Q2, and Q3 to Q2.
F12 = kQ1*Q2/r² = (9 * 10^9 * 8 * -5 * 10^-12)/(0.05)² = -360 * 10^-3/0.0025 = -144 N
F32 = kQ3*Q2/r² = (9 * 10^9 * 30* -5 * 10^-12)/(0.1)² = - 1350 * 10^-2/0.01 = - 1350 N
So, net force on Q2 is,
F = F12 – F32 = -144 - (-1350) = -144 + 1350 = 1206 N
6.
Referring to the diagram below.
What is the magnitude and direction of the resultant force
acting on Q2?
Answer: Using the same concept, magnitude of the net force on Q2 is,
F12 = kQ1*Q2/r² = (9 * 10^9 * 8 * -6 * 10^-12)/(0.03)² = (-48 * 10^-3 * 9 * 10^4)/9 = -480 N
Similarly, F32 = kQ3*Q2/r² = (9 * 10^9 * 8 * -6 * 10^-12)/(0.04)² = -270 N.
Since, both forces are directed perpendicular to each other, net force is addition of the two of them,
i.e.
-480 – 270 = - 750 N.
Direction of the force is, theta = tan^-1(-480/-270) = tan^-1(1.778) = 60.64 degrees..
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