How much work does it take for an external agent to move a 45.0-nC charge from a point on the +x-axis, 3.40 cm from the origin t
o a point halfway between the 41.0-nC and 52.0-nC charges?
1 answer:
Answer:
The work done on the 45.0nC charge is 2.24×10^-3J. The detailed solution can be found in the attachment below.
Explanation:
The Problem solution makes use of the potential energy relationship between the charges. This solution assumes the distance as shown in the diagram. For a different distance arrangement adjustments should be made appropriately.
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Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.
3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.