A ring of charge is centered at the origin in the vertical direction. The ring h

Question

A ring of charge is centered at the origin in the vertical direction. The ring has a charge density of lambda = 6. 93 times 10-6 C/m and a radius of R = 1. 73 cm. Find the total electric field, E, of the ring at the point P = (1. 79 m, 0. 00 m). The Coulomb force constant is k = 1/(4piepsilon0) = 8. 99 times 109 N m2/C2. E = Find the expression for the electric field, Einfinity, of the ring as the point P becomes very far from the ring (x R) in terms of the radius R, the distance x,the total charge on the ring q, and the constant k = 1/(4piepsilon0). Einfinity =

Explanation / Answer

1) You can break this problem up into three parts. First, when the block begins the accelerate towards the wall, then when the block hits the wall, and finally when it beings to move in the opposite direction but decelerating. For the first part, you know the only force being applied is a uniform electric field to the right. Thus, F(electric) = ma. Since you know the mass, and you can calculate the force of the electric field (you know d and q) and use this in conjunction with m to solve for the acceleration. You can then use the equations of linear momentum to find the time it takes to hit the wall (specifically, you can use d = d_i + (v_i)(t) + (1/2)(a)(t²), where d_i = v_i = 0). Since you're assuming the collision is perfectly elastic, all the force is transferred into the wall, and back into the object. Thus, the kinetic energy the block at the point of collision will be reversed in direction, but have the same magnitude. For the last part, it's carrying a constant velocity at the point of collision, and is now traveling towards the left. The electric field will be causing it to decelerate. Assuming no friction and no other energy loss, it should return to the point it started, and being the cycle all over again. Using this information, you can solve for the period of oscillation, from the equation of linear motion, as the time should be equal for the return trip. 2)There's few potential positions for this question. Firstly, the third particle must be placed on the same plane as the other two particles, else force will be split into several components for at least one of the particles, which will reduce its strength. You can't put it between the two particles, as the two forces will be pushing the third particle in opposite directions, limiting the net force. Thus, there's only two possible scenarios. Assuming the setup is, (a)---------d---------(b) (distance d between) you can either put the third particle to the left of a or to the right of b. Since the farther away the particle is, the less force it experiences, it has to be as close as possible to either particle without going in between the two. Thus, it has to be directly beside a (to the left of it) or directly beside b(to the right of it) to experience the maximum force, assuming like charges. 3) The great thing about rings, is that if you put anything at the center, the electric field is 0 due to equal and opposite forces being applied to a charge from every direction. Now, if snip small sections off, the only charges contributing to the force in the center is the small section of ring on the opposite end of the ring, as the rest of the ring will still be nullified due to opposite and equal competing forces in the ring center. In general, you can use the equation, ?dE = ?(1 / 4­pe)(dq / r) = ?(? / 4­pe)(ds / r), as dq = ?ds, where ds is the arc length of a differential element of remaining wire and ? is the charge distribution (Q over the length if wire). The only section you would need to integrate over is the section of wire opposite to the missing wire section (as the rest of the wire would have a net force of 0). However, if you take this section of wire and split into separate x and y components, you may find that if you integrate from the center outward in each direction that one component will cancel and the other will have charges of equal magnitude and in the same direction toward the middle of this wire section. Thus, the electric field would be E = Ex = (dE)cosß, noting that s = rß and ds = r(dß) (assuming you set up the axis such that its the y components that cancel). Note that r is the radius of the ring, s is the arc length or a section of ring, and ß is the angle of said arc length. You can either integrate over the arc length, or convert it to angles using ds = r(dß) and integrate that way. 4) This problem seems pretty straight forward. You need to draw a free-body diagram of the forces being applied to the bob, and find how the forces are in equilibrium (Fnet = Fb - Fa = 0). There's only three forces from the looks of it, which are tension, gravity (mg) and force due to the electric field on the charged bob. However, it would be difficult for me to go into any more detail unless I know the direction of the electric field, and what angles are made from the thread. But, I'd put the axis over the bob such that it's at the origin, and work from there.

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