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Problem 14 Medium Difficulty

Two identical metal blocks resting on a frictionless horizontal surface are connected by a light
metal spring having constant $k=100 \mathrm{N} / \mathrm{m}$ and unstretetehed length $L_{i}=0.400 \mathrm{m}$ as in Figure $\mathrm{P} 15.14 \mathrm{a}$ . A charge $Q$ is slowly placed on each block causing the spring to stretch to an cquilibrium length $L=0.500 \mathrm{m}$ as in Figure P15.14b. Determinc the value of $Q$ modeling the blocks as charged particles.

Answer

$1.67 \times 10^{-5} \mathrm{C}$

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Video Transcript

so there's two forces acting on each block. One of the forces is the force of spring, which is pulling the two blocks closer together. On the second force is the force between the two charges which is pulling the blocks further apart because the blocks are not moving into an equilibrium. We know that the two forces are equal, so we can say that the force of the spring is equal to the force between the two charges. So what we're gonna do is we're gonna look at the equation for both the force of the spring and force of the charge and use that to solve for Q. So looking at the force of the spring, first we get an equation from hooks lot, which is the spring constant K multiplied by the displacement of the spring. In this problem, K is given as 100 Newtons per meter and X is 0.1 meter. The displacement. The 0.1 meter is found by taking the difference between the spring length at equilibrium and the spring length when it's resting. And so with this equation we find that the forces of spring is equal to 10 Newtons, so from that. We also know that the force between the two charges is equal to 10 should now a solve. We'll find the equation for force between two charges. And so this equation is given by Coolum slaw, and we know that to be ah, universal constant K multiplied by the value of the two charges Q wanted you to and divided by the distance between them squared, which is our square K's universal constant. And we know that to be 8.99 times 10 to the negative night Newton's times meter square for cool. The two charge values are not known, and that's what we're looking for. But we do know that the two values are the same. So that's cute times Q. Which we can represent its cute square and then in the denominator, you have the distance between the charges, which is 0.5 meters square, and so here we have a fraction for the force between the charges, which we know to be 10 noon. So now we will just solve that 10 Newtons is equal to this value on the right, and I'm gonna rewrite it. But I'll drop unit so it's easier to see. And so all that stuff to do is to solve it. So multiply. Resolving for Q. Here. So we multiply both sides by 0.5 squared and on the right that cancels with nominator. I will divide both sides by 8.99 times 10 to the negative night and on the right that cancels with the numerator. So on the right side, you'll notice all that's left is Q Square, which is great because that's what we're solving for. So all that's left is to take the square root of both sides and on the right side you have cute on the left side, you have value that when you solve, you get 1.67 times 10 to the negative. Fifth cools, and this is the solution we're looking for. And so one thing to note about this problem is that the charges repel each other. So that means either both charges air positive or both are negative, but either answer would be correct.

University of Minnesota - Twin Cities
Top Physics 102 Electricity and Magnetism Educators
Andy C.

University of Michigan - Ann Arbor

Marshall S.

University of Washington

Zachary M.

Hope College

Aspen F.

University of Sheffield

Physics 102 Electricity and Magnetism Bootcamp

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