What would the electrostatic force be for two 0.005C charges 50m apart?
1 answer:
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Answer:
v_average = 15 m / s
Explanation:
The average speed can be found in two ways,
* taking the distance traveled and divide it by the time spent
* taking the velocities in each time interval and then finding the weighted average by the time fraction
v_average = 1 / t_total ∑ vi ti
Let's apply this last equation
Total time is
t = t₁ + t₂
t = 10 + 10 = 20 min
v_average = 10/20 10 + 10/20 20
v_average = 10/2 + 20/2
v_average = 15 m / s
The reasoning is wrong if we look into Newton's Law of gravitation.
Newton's law of gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
The law is written as follows;
The distance between the two particles, is a function of force and their masses not necessarily time of motion.
In the given problem only time of motion was considered which is wrong.
Thus, the reasoning is wrong if we look into Newton's Law of gravitation.
Learn more here: brainly.com/question/19680441
Answer:
Explanation:
Plate separation, d = 1.76 cm = 0.0176 m
Area of plates, A = 25 cm^2 = 0.0025 m^2
V = 255 V
(a) Capacitance of capacitor
C = 1.258 x 10^-12 F
charge is same before and after immersion as the battery is disconnected
q = C V
q = 1.258 x 10^-12 x 255 = 3.2 x 10^-10 C
(b)
Capacitance before, C = 1.258 x 10^-12 C
capacitance after, C' = k x C = 80 x 1.258 x 10^-12 = 100.64 x 10^-12 C
Where, k is the dielectric constant of water = 80
Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V
(c) initial energy,
Final energy