Fall 2011 Homework 6 Solutions

# Fall 2011 Homework 6 Solutions - Physics 21 Fall 2011...

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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-6 23-9 A point charge q 1 = 4 . 10 nC is placed at the origin, and a second point charge q 2 =- 2 . 9 nC is placed on the x-axis at x = +21 . 0 cm. A third point charge q 3 = 2 . 1 nC is to be placed on the x-axis between q 1 and q 2 . (Take as zero the potential energy of the three charges when they are infinitely far apart.) a) What is the potential energy of the system of the three charges if q 3 is placed at x = +11 . 0 cm? b) Where should q 3 be placed to make the potential energy of the system equal to zero? (a) The general formula for the potential energy between any two point charges labeled 1 and 2 is U 12 = 1 4 πǫ q 1 q 2 | r 1- r 2 | , where | r 1- r 2 | is the distance between the two charges. To find the total potential energy of several charges, we must find the potential energy due to each pair of point charges. So the total potential energy here is U total = U 12 + U 13 + U 23 = 1 4 πǫ parenleftbigg q 1 q 2 |- . 21 m | + q 1 q 3 |- . 11 m | + q 2 q 3 | . 21 m- . 11 m | parenrightbigg =- 3 . 53 × 10- 7 J (b) We can say that q 3 is placed at some point x between the other two charges and determine which value of x gives the system a total potential energy of zero. Incorporating this into the expression for the total potential energy gives: U total = 1 4 πǫ parenleftbigg q 1 q 2 . 21 m + q 1 q 3 x + q 2 q 3 ( . 21 m- x ) parenrightbigg = 0 We can save ourselves some trouble by eliminating several overall factors that won’t affect the solution x . Since all the terms have exactly the same units, we can drop the conver- sion factors to SI units and write distances in cm and charges in nC. The result is (4 . 1)(- 2 . 9) 21 + (4 . 1)(2 . 1) x + (- 2 . 9)(2 . 1) (21- x ) = 0 Multiply through by x (21- x ) to get a quadratic equation: . 5662 x 2- 26 . 59 x + 180 . 81 = 0 Using the quadratic formula we find x = 38 . 72 or 8.25. Re- member that these values are in centimeters. Since the firstmember that these values are in centimeters....
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Fall 2011 Homework 6 Solutions - Physics 21 Fall 2011...

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