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2 years ago

Two positive charges are equal. Which has more electric potential energy?

Physics

2 answers:

Maurinko [17]2 years ago

8 0

<h2>Answer 1 with Explanation </h2>

Electric potential vitality is the vitality that is expected to move a charge against an electric field. You need more vitality to move a charge further in the electric field, yet in addition more vitality to move it through a more grounded electric field. Envision that you have a colossal contrarily charged plate, with a little emphatically charged molecule adhered to it through the electric power. There's an electric field around the plate that is pulling all emphatically charged items toward it (while pushing other adversely charged articles away). You take the positive molecule, and begin to pull it off the plate, against the draw of the electric field. It's diligent work, on the grounds that the electric power is pulling them together.

<h2>Answer 2 with Explanation </h2>

On the off chance that you let the positive molecule go, it would snap back to the negative plate, pulled by the electric power. The vitality that you used to move the molecule far from the plate is put away in the molecule as electrical potential vitality. The potential the molecule needs to move when it's given up. In the event that you pulled the positive molecule further far from the plate, you would need to utilize more vitality, so the charge would have increasingly electrical potential vitality put away in it. In the event that we multiplied the charge on the plate, once more, you would require more vitality to move the positive molecule. On the off chance that we multiplied the charge on the positive molecule, you would require more vitality to move it. You get the thought. Envision that rather than an adversely charged plate, our plate is decidedly charged. Our positive molecule would be pushed far from the plate since they are both decidedly charged. This time, we need to put in vitality to endeavor to draw the molecule nearer to the plate, rather than to pull it away. The closer we endeavor to move it to the plate, the more vitality we need to put in, so the more electrical potential energy the particle would have.

Xelga [282]2 years ago

7 0

Electrostatic potential energy of a system of charge is given by

$U = \frac{kq_1q_2}{r}$

here we have

$q_1,q_2$ = two charges of different magnitudes

r = distance between charges

so here we can see that electrostatic potential energy will depends upon the product of two charges and inversely depends upon the distance between the two charges

So here we can say that the electrostatic potential energy of two charges will be same and equal to each other

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We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

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Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

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$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

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