Fall 2011 Homework 3 Solutions

Fall 2011 Homework 3 Solutions - Physics 21 Fall, 2011...

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Physics 21 Fall, 2011 Solution to HW-3 21-96 Positive charge Q is uniformly distributed around a semicircle of radius a . Find the electric ±eld (magnitude and direction) at the center of curvature P . x y d θ dQ = λ ds = λ ad θ We start with the equation from the equation sheet that gives the ±eld at the ±eld point r in terms of a charge dQ at the charge point r : d E ( r ) = 1 4 πǫ 0 dQ ( r - r ) | r - r | 3 The ±eld point r (where we want to know E ) is at the origin, so r = 0, and the charge point (where dQ is located) is r = a cos θ ˆ i + a sin θ ˆ j , where θ is the angle above the x -axis, so r - r = - a cos θ ˆ i - a sin θ ˆ j , and | r - r | = a. Since we want to add up all the contribution from all dQ , we must relate dQ to so that we can integrate over the angle θ spanned by the semicircle. The arc length ds swept out by an angle is adθ , so the charge dQ on ds is dQ = λds = λa dθ, where λ is the linear charge density (charge per unit length): λ = Q πa . With all these substitutions, the original equation becomes d E = 1 4 πǫ 0 ( Q πa adθ )( - a cos θ ˆ i - a sin θ ˆ j ) a 3 = - Q 4 π 2 ǫ 0 a 2 (cos θ ˆ i + sin θ ˆ j ) dθ. We can ±nd the total ±eld at the origin by integrating: E = i d E = i π 0 - Q 4 π 2 ǫ 0 a 2 (cos θ ˆ i + sin θ ˆ j ) dθ. We will look at each component of the integral over θ sepa- rately. We see that i π 0 cos θ dθ = 0 and i π 0 sin θ dθ = 2 , so we get E = - Q 4 π 2 ǫ 0 a 2 (0 ˆ i + 2 ˆ j ) = - Q 2 π 2 ǫ 0 a 2 ˆ j . From this result we see the E has a magnitude of Q 2 π 2 ǫ 0 a 2 and points in the - y direction or downward. The x compo- nent of the ±eld is zero, as one would expect from symmetry. 21-105 Three charges are placed as shown in the ±gure. The magnitude of q 1 is 2.00 µ C, but its sign and the value of the charge q 2 are not known. Charge q 3 is +4.00 µ C, and the net force F on q 3 is entirely in the negative x -direction. a) Calculate the magnitude of q 2 . b) Determine the magnitude of the net force F on q 3 . (a) We ±rst determine the sign of the charge q 1 . We can do this by thinking about which direction the force will be in for the di²erent combinations of signs for charges q 1 and q 2 . Since there is no y -component of the force on q 3 we know that q 1 and q 2 must have opposite signs. Since the force is directed in the negative x -direction we can infer that q 1 must be negative and q 2 must be positive. To determine q 2 we calculate the the total force F = F 1 on 3 + F 2 on 3 on q 3 . Expressions for F 1 on 3 and F 2 on 3 follow from the general expression for F 2 on 1 on the equation sheet: F 1 on 3 = 1 4 πǫ 0 q 3 q 1 ( r 3 - r 1 ) | r 3 - r 1 | 3 , F 2 on 3 1 4 πǫ 0 q 3 q 2 ( r 3 - r 2 ) | r 3 - r 2 | 3 . We need the position vectors of each charge. Let
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This note was uploaded on 12/18/2011 for the course PHYS 021 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Fall 2011 Homework 3 Solutions - Physics 21 Fall, 2011...

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