A point charge Q=+4.60 μC is held fixed at the origin. A second point charge q=+1.20 μC with mas

A point charge Q=+4.60 μC is held fixed at the origin. A...

A point charge $Q=+4.60 \mu \mathrm{C}$ is held fixed at the origin. A second point charge $q=+1.20 \mu \mathrm{C}$ with mass of $2.80 \times$ $10^{-4} \mathrm{kg}$ is placed on the $x$ axis, 0.250 $\mathrm{m}$ from the origin. (a) What is the electric potential energy $U$ of the pair of charges? (Take $U$ to be zero when the charges have infinite separation.) (b) The second point charge is released from rest. What is its speed when its distance from the origin is (i) $0.500 \mathrm{m} ;$ (ii) 5.00 $\mathrm{m}$ ; (iii) 50.0 $\mathrm{m} ?$

Key Concepts

Electric Potential Energy

This concept refers to the energy stored in a system of charges due to their positions relative to one another. For point charges, the electric potential energy is typically defined as U = kq1q2/r, where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. This energy represents the work that would be required to bring the charges from infinite separation to the specified distance, and it is taken as zero when the charges are infinitely apart.

Electric Potential

Electric potential is defined as the potential energy per unit charge at a point in an electric field, and for a point charge it is given by V = kQ/r. This allows us to calculate the potential energy of any test charge placed in the field by simply multiplying the test charge by the electric potential. It provides a scalar description of the influence of the electric field at various points in space.

Conservation of Energy in Electrostatics

This key concept implies that in the absence of non-conservative forces, the total mechanical energy (the sum of kinetic and potential energies) of a system remains constant. When charges move under the influence of electric forces, any decrease in electric potential energy corresponds to an increase in kinetic energy. This principle is essential for solving problems where a charge is released and accelerates, allowing one to determine its speed at different positions by equating the loss in potential energy to the gain in kinetic energy.

Transcript

00:01 Okay, so in this problem, we're told that a point charge is at the origin.

00:06 So here's q.

00:08 Q is plus 4 .60 times 10 to the minus 6 kuloms.

00:18 And here it is at the origin.

00:19 And then the second one is placed 0 .25 at a position of 0 .25 from the origin.

00:27 So that's this q and that's 1 .2o microchulomes.

00:37 And then this particle also has a mass of 2 .8 -0 times 10 to the minus 4 kilograms.

00:49 So here's our q and then our other q.

01:05 And then we were trying to obtain the electric potential energy of the pair.

01:11 So that's just going to be, this is probably a formula in the book, kqq over r.

01:21 So i'm going to plug that into a calculator.

01:25 I obtained 0 .199 joules.

01:39 And for part b, we're told that the second particle is released from rest, and we were told that the second particle is released from rest.

01:45 Want to know the speed when it's at these different distances from the origin.

01:52 So i'll just go ahead and record those distances.

01:56 So b, our distances are 0 .5 meters.

02:05 Are these opposite sign charge charges? oh, yeah, they're the same size.

02:16 Okay, so they'll repel.

02:17 So 0 .5 and then five and then 50.

02:24 Oops, one, two, and then three.

02:33 So for this, we want to use that the chain, if there's no work being, external work being done on the system, then the change in kinetic energy is equal to minus the change in potential energy by energy conservation...

Please give Ace some feedback