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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman
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Positive Charge +Q is distributed uniformly along the

Chapter 21, Problem 95P

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QUESTION:

Positive Charge +Q is distributed uniformly along the +x-axis from x = 0 to x = a. Negative charge ?Q is distributed uniformly along the ?x-axis from x = 0 to x = ?a. (a) A positive point charge q lies on the positive y-axis, a distance y form the origin. Find the force (magnitude and direction) that the positive negative charge distributions together exert on q. Show that this force is proportional to y?3 for y ? a. (b) Suppose instead that the positive point charge q lies on the positive x-axis, a distance x > a from the orgin. Find the force (magnitude and direction) that the charge distribution exerts on q. Show that this force is proportional to x?3 for x ? a.

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QUESTION:

Positive Charge +Q is distributed uniformly along the +x-axis from x = 0 to x = a. Negative charge ?Q is distributed uniformly along the ?x-axis from x = 0 to x = ?a. (a) A positive point charge q lies on the positive y-axis, a distance y form the origin. Find the force (magnitude and direction) that the positive negative charge distributions together exert on q. Show that this force is proportional to y?3 for y ? a. (b) Suppose instead that the positive point charge q lies on the positive x-axis, a distance x > a from the orgin. Find the force (magnitude and direction) that the charge distribution exerts on q. Show that this force is proportional to x?3 for x ? a.

ANSWER:

Solution 95P Introduction We have to calculate the total force on the charge q due to the charge distribution when the charge q is on the y axis and on the x axis. Step 1: Because of the symmetry of the problem, the y component of the force for the negative and positive part will be canceled out and only the x component will survive. Hnce we will calculate the x component for each distribution and then add them. Also the because of the same symmetry the x component of the force will be same for both. Hence we will calculate the x component once then we can just multiply by two. To calculate the force, let us consider an elementary length of charge dx at a distance x fro the origin, and then we will calculate the force due to that elementary length and then we will integrate over the whole range, that is from a to 0. The linear charge density of the charge distribution of the negative side is Hence the charge of the elementary length is Hence the force due to the part to the point is Where is the distance of the charge from the elementary length. And . Hence the component of the force is Now putting all the values we have Hence the total x component of the force is So the total component of the force due to the both side of the charge distribution will be And the direction of the force is towards the negative axis.

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