Answer:
(A)
Explanation:
We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.
Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.
In theory, yes. The 2 problems are the materials used for clinical thermometers, & the temperature capacity of the clinical thermometer. If anything, change the material & extend the measurement threshold. At that point, it wouldn´t be used for clinical garbage anymore.
We are in the lowest layer called Troposphere.
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The charge present determines a force to be attractive or repulsive.
The charges acquired by two bodies determines the Force as Attractive Or Repulsive.
Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.
Examples of the experiments and observations:
- On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.
This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.
- In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.
The more the distance between the charges, the less is the Electric Force.
The lesser the distance between the charges, the more is the Electric Force.
If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.
Hence, the charge present determines a force to be attractive or repulsive.
Learn more about Coulomb Force here, brainly.com/question/15451944
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Answer:
The electric potential will be "259.695 volt".
Explanation:
In the given question, the figure is not provided. Below is the attached figure given.
Given:
Now,
At point P, the electric potential will be:
⇒ 
By putting values, we get
⇒ ![=9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]](https://tex.z-dn.net/?f=%3D9%5Ctimes%2010%5E9%20%5B%5Cfrac%7B6.39%5Ctimes%2010%5E%7B-9%7D%7D%7B0.40%7D%20%2B%5Cfrac%7B3.22%5Ctimes%2010%5E%7B-9%7D%7D%7B0.25%7D%20%5D)
⇒ 
