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A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a\), and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{~J}\). When the second charge is moved to point \(b,\) the clectric force on the charge does \(-1.9 \times 10^{-8} \mathrm{~J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b ?\)

Short Answer

Expert verified
The electric potential energy of the pair of charges when the second charge is at point \(b\) is \(3.5 \times 10^{-8}\) Joules.

Step by step solution

01

Understand and note the given values

The initial potential energy \(U_a\) when the second charge is at point \(a\) is given as \(+5.4 \times 10^{-8}\) Joules. The work \(W\) done by the electric force when the charge moves from \(a\) to \(b\) is \(-1.9 \times 10^{-8}\) Joules.
02

Understand the concept of conservation of energy

According to the principle of conservation of energy, the total mechanical energy remains constant in the absence of non-conservative forces. In this case, the initial potential energy \(U_a\) plus the work done by the electric force \(W\) should equal the final potential energy \(U_b\) at point \(b\). So we have \(U_a + W = U_b\).
03

Substitute the given values and compute for \(U_b\)

Substitute the given values \(U_a = 5.4 \times 10^{-8}\) Joules and \(W = -1.9 \times 10^{-8}\) Joules into the equation to find \(U_b\). Solving \(5.4 \times 10^{-8} - 1.9 \times 10^{-8} = U_b\) results in \(U_b = 3.5 \times 10^{-8}\) Joules. So, the electric potential energy of the pair of charges when the second charge is at point \(b\) is \(3.5 \times 10^{-8}\) Joules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
When discussing electric potential energy, the conservation of energy is a fundamental concept that plays a crucial role. It states that the total energy in an isolated system remains constant over time. This means that energy can neither be created nor destroyed; it can only be transformed from one form to another or transferred from one object to another.

In the context of point charges, as seen in the exercise, conservation of energy implies that the electric potential energy combined with any work done by or against electric forces results in a new form of energy or a redistribution of energy within the system. If we have a charge in an electric field and move it from one point to another, the work done on the charge changes its electric potential energy.

For our point charges example, this can be observed when the charge is moved from point a to point b. The electric potential energy at point a and the work done to move the charge to point b together must equal the electric potential energy at point b, assuming no other forces are doing work. This concept ensures that the initial and the final state of the system can be connected through energy transformations without loss or gain, adhering to the conservation of energy principle.
Work-Energy Principle
The work-energy principle is another pivotal concept to understand electric potential energy in a physical context. This principle tells us that the work done on an object is equal to the change in its kinetic energy. However, when dealing with point charges in an electric field, the principle extends to relate work done to the change in electric potential energy, especially when no kinetic energy is involved, as charges can be held stationary.

Applying this principle to our exercise, the work done by the electric force when the charge moves from point a to b is negative because the force opposes the displacement. This negative work indicates that energy is being taken away from the charge, thereby reducing its electric potential energy. To calculate the final electric potential energy at point b, we can deduct the work done from the initial electric potential energy at point a.

This aligns with our intuition that if you have to 'work against' a force to move an object, you're decreasing its potential in the field created by that force. Since the scenario provided does not include kinetic energy, the focus lies solely on this transfer between work and electric potential energy.
Point Charges
The term point charges refers to hypothetical charges that are concentrated at a single point in space with no volume. Despite this being an idealization, it serves as an extremely useful concept in understanding electric forces and fields.

In practice, when considering point charges like charges q_{1} and q_{2} in our exercise, the electric potential energy of the system depends on the relative positions of these charges as well as their magnitudes. The closer the charges are, the stronger the interaction and the higher the potential energy if they are of opposite signs, whereas charges of same sign would have higher potential energy when separated.

Calculating Electric Potential Energy

To calculate the electric potential energy between point charges, you can use Coulomb's law, which relates the potential energy to the product of the charges divided by the distance between them and multiplied by the Coulomb constant. However, in our exercise, we already have the initial electric potential energy and the work done, so we do not need to calculate this from scratch, but rather apply conservation of energy to find the new electric potential energy after charge q_{2} moves to a new location.

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Most popular questions from this chapter

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

A point charge \(+8.00 \mathrm{nC}\) is on the \(-x\) -axis at \(x=-0.200 \mathrm{~m}\) and a point charge \(-4.00 \mathrm{nC}\) is on the \(+x\) -axis at \(x=0.200 \mathrm{~m}\). (a) In addition to \(x=\pm \infty\), at what point on the \(x\) -axis is the resultant field of the two charges equal to zero? (b) Let \(V=0\) at \(x=\pm \infty\), At what two other points on the \(x\) -axis is the total electric potential due to the two charges equal to zero? (c) Is \(E=0\) at either of the points in part (b) where \(V=0 ?\) Explain.

Points \(A\) and \(B\) lie within a region of space where there is a uniform electric field that has no \(x\) - or \(z\) -component; only the \(y\) -component \(E_{y}\) is nonzero. Point \(A\) is at \(y=8.00 \mathrm{~cm}\) and point \(B\) is at \(y=15.0 \mathrm{~cm} .\) The potential difference between \(B\) and \(A\) is \(V_{B}-V_{A}=+12.0 \mathrm{~V},\) so point \(B\) is at higher potential than point \(A\). (a) Is \(E_{y}\) positive or negative? (b) What is the magnitude of the electric field? (c) Point \(C\) has coordinates \(x=5.00 \mathrm{~cm}, y=5.00 \mathrm{~cm} .\) What is the potential difference between points \(B\) and \(C ?\)

CP Deflecting Plates of an Oscilloscope. The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying cqual but opposite charges. Typical dimensions are about \(3.0 \mathrm{~cm}\) on a side, with a separation of about \(5.0 \mathrm{~mm} .\) The potential difference between the plates is \(25.0 \mathrm{~V}\). The plates are close enough that we can ignore fringing at the cnds. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

An object with charge \(q=-6.00 \times 10^{-9} \mathrm{C}\) is placed in a region of uniform electric field and is released from rest at point \(A .\) After the charge has moved to point \(B, 0.500 \mathrm{~m}\) to the right, it has kinetic energy \(3.00 \times 10^{-7} \mathrm{~J}\). (a) If the electric potential at point \(A\) is \(+30.0 \mathrm{~V}\). what is the clectric potential at point \(B ?\) (b) What are the magnitude and direction of the electric field?

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