Fall 2011 HW 2 Solutions

# Fall 2011 HW 2 Solutions - Physics 21 Fall 2011 Solution to...

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Physics 21 Fall, 2011 Solution to HW-2 21-13 Three point charges are arranged on a line. Charge q 3 = +5 . 00 nC and is at the origin. Charge q 2 = 3 . 00 nC and is at x 2 = 4 . 50 cm. Charge q 1 is at x 1 = 1 . 00 cm. What is q 1 (magnitude and sign) if the net force on q 3 is zero? x 3 =0 x 1 x 2 q 3 q 1 q 2 We can work this problem without using vectors by think- ing it through. Since q 2 and q 3 have opposite sign, the force on q 3 exerted by q 2 is attractive (towards the right). If the total force on q 3 is to be zero, the force exerted by q 1 must be repulsive (toward the left). Thus q 1 and q 3 must have the same sign, and q 1 must be positive. We can find the magnitude of q 1 by equating the magni- tude of the forces on q 3 exerted by q 1 and q 2 : 1 4 π 0 | q 1 q 3 | x 2 1 = 1 4 π 0 | q 2 q 3 | x 2 2 Cancelling like terms on both sides of the equation, we find | q 1 | = x 1 x 2 2 | q 2 | = 1 . 0 cm 4 . 5 cm 2 (3 nC) = 0 . 148 nC . We already concluded that q 1 was positive. A more general way to solve this problem is to use the vector expressions for the Coulomb force. We want 0 = F 1 on 3 + F 2 on 3 , where 0 = 1 4 π 0 q 1 q 3 ( r 3 r 1 ) | r 3 r 1 | 3 + 1 4 π 0 q 2 q 3 ( r 3 r 1 ) | r 3 r 2 | 3 . and we can evaluate the forces using the locations of the charges. Because q 1 is at the origin, r 3 = 0. Also, r 1 = x 1 ˆ i and r 2 = x 2 ˆ i . Substituting for the vectors gives 0 = 1 4 π 0 q 1 q 3 (0 x 1 ) ˆ i | 0 x 1 | 3 + q 2 q 3 (0 x 2 ) ˆ i | x 2 | 3 = q 3 4 π 0 q 1 x 1 | x 1 | 2 + q 2 x 2 | x 2 | 2 ˆ i

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