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Consider two identical conductor spheres, A
and B. Initially, sphere A has a charge of
-80Q and sphere B has a charge of +40Q. If
the spheres touched and then are separated
by a distance of 0.4m. What is the resultant
force between them?(Q=6uC)
Select one:
OON
1.8X10%N, attraction
8.1X102N, attraction
1.8X10°N, repulsive
3.2X10ⓇN, attraction
3.2X10°N, repulsion
8.1X10N, repulsive

2 Answers

5 votes

Answer:

1.08 ×10^9 N attraction

Step-by-step explanation:

By employing Columbs law of force between charge .the force is defined as;

F = KQ1Q2/ r2

= 9×10^9 ×80×40×(6×10^-6) / 0.4 ^2

= 1.08 ×10^9 N

K = 9×10^9 N m2/C2

User Jagmal
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5.6k points
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The answer is the the first one
User Tom Elliott
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5.2k points