ch23-p060 - ⎠ ⎟ ⎞ q 4 πε o R 2 r = 1 2 e q 4 πε o...

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where r is measured from the center of the ball (to the proton).This agrees with Coulomb’s law from Chapter 22. We note that if r = R then this expression becomes F R = e q 4 πε o R 2 . (a) If we require F = 1 2 F R , and solve for r , we obtain r = 2 R . Since the problem asks for the measurement from the surface then the answer is 2 R R = 0.41 R . (b) Now we require F inside = 1 2 F R where F inside = eE inside and E inside is given by Eq. 23-20. Thus, e
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Unformatted text preview: ⎠ ⎟ ⎞ q 4 πε o R 2 r = 1 2 e q 4 πε o R 2 ⇒ r = 1 2 R = 0.50 R . 60. The field at the proton’s location (but not caused by the proton) has magnitude E . The proton’s charge is e . The ball’s charge has magnitude q . Thus, as long as the proton is at r ≥ R then the force on the proton (caused by the ball) has magnitude F = eE = e ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ q 4 πε o r 2 = e q 4 πε o r 2...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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