The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
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Electric potential energy</h3>
When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.
The electric potential energy between the charges when the second charge is at point b is calculated as follows;
ΔU = -w
Ui - Uf = w
Uf = Ui - w
where;
Uf is the final potential energy
Ui is the initial potential energy
w is the work done by the force
Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)
Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J
Uf = 7.3 x 10⁻⁸ J
Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
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A piece of purple plastic is charged with 3. 13×106 extra electrons compared to its neutral state, then the net electric charge in coulomb would be - 5.008×10⁻¹³ coulombs.
<h3>What is an electric charge?</h3>
Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge (commonly carried by protons and electrons respectively). Unlike charges attract one another while like charges repel one another. We refer to an object as neutral if it has no net charge.
The charge on one electron is -1.6 ×10⁻¹⁹ coulomb.
Then the charge on the 3.13×10⁶ extra electrons compared to its neutral state
=-1.6×10⁻¹⁹ ×(3.13×10⁶)
As given in the problem A piece of purple plastic is charged with 3.13×10⁶ extra electrons compared to its neutral state then the net electric charge in coulombs would be - 5.008×10⁻¹³ coulombs.
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We know that when you heat up calcium sulfate all the water has evaporated from it. That means that the mass of water that was initially in the sulfate is the difference in mass of the sulfate before and after the heating.
In order to <span>get a percentage by mass we simply divide this number by the initial amount of sulfate:
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Explanation:
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