Q = magnitude of charge at each of the three locations A, B and C = 2 x 10⁻⁶ C
r₁ = distance of charge at origin from charge at B = 50 - 0 = 50 cm = 0.50 m
r₂ = distance of charge at origin from charge at C = 100 - 0 = 100 cm = 1 m
F₁ = magnitude of force by charge at B on charge at origin
F₂ = magnitude of force by charge at C on charge at origin
Magnitude of force by charge at B on charge at origin

inserting the values

F₁ = 0.144 N
Magnitude of force by charge at C on charge at origin

inserting the values

F₂ = 0.036 N
Net force on the charge at the origin is given as
F = F₁ + F₂
F = 0.144 + 0.036
F = 0.18 N
from the diagram , direction of net force is towards left or negative x-direction.
Answer:
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Explanation:
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It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway, here is the answer. The revival of the atomic concept was catalyzed by the experimental observation which is Constant <span>composition in compounds. Hope this helps.</span>
Work = force × distance × cos(angle)
work = (25)(50)(cos (40))
work = 957.56 Joules
= 9.6x10^2 Joules
The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,

where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?




Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
Learn more about capacitive reactance here:
brainly.com/question/23427243
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