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Chapter 20: Problem 10

Under what circumstances is the path of a charged particle a parabola? A circle?

Short Answer

Expert verified
A charged particle will move in a parabolic path when it's under the influence of a uniform electric field, and will move in a circular path when it enters a uniform magnetic field perpendicular to its direction of motion.

Step by step solution

01

Determining the Path of Parabolic Motion

A path is parabolic when a force acting on the particle is constant and in one direction. For a charged particle, this occurs when it's under the influence of a uniform electric field. The constant electrical force \( F_e = qE \) (where \( q \) is the charge of the particle and \( E \) is the electric field strength) would cause the particle to accelerate uniformly, resulting in parabolic motion.
02

Determining the Path of Circular Motion

A charged particle will move in a circular path when it enters a uniform magnetic field perpendicular to its velocity. In such a case, the only force acting on the particle is the magnetic force \( F_m = qvB \) (where \( v \) is the velocity of the particle and \( B \) is the magnetic field strength). Since this force is perpendicular to the motion and depends on the velocity, it results in the particle moving along a circular path.

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Most popular questions from this chapter

In his famous 1909 experiment that demonstrated quantization of electric charge, R. A. Millikan suspended small oil drops in an electric field. With field strength \(20 \mathrm{MN} / \mathrm{C},\) what mass drop can be suspended when the drop carries 10 elementary charges?

Two identical small metal spheres initially carry charges \(q_{1}\) and \(q_{2} .\) When they're \(1.0 \mathrm{m}\) apart, they experience a \(2.5-\mathrm{N}\) attractive force. Then they're brought together so charge moves from one to the other until they have the same net charge. They're again placed \(1.0 \mathrm{m}\) apart, and now they repel with a \(2.5-\mathrm{N}\) force. What were the original charges \(q_{1}\) and \(q_{2} ?\)

Equation 20.3 gives the electric field of a point charge. Does the direction of (a) \hat{r } or (b) \(\vec{E}\) depend on whether the charge is positive or negative?

Why should there be a force between two dipoles, which each have zero net charge?

A \(65-\mu C\) point charge is at the origin. Find the electric field at the points (a) \(x=50 \mathrm{cm}, y=0 \mathrm{cm} ;\) (b) \(x=50 \mathrm{cm}, y=50 \mathrm{cm}\) and (c) \(x=25 \mathrm{cm}, y=-75 \mathrm{cm}\)
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