oldhomework 01-solutions

# oldhomework 01-solutions - howard(cah3459 oldhomework 01...

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howard (cah3459) – oldhomework 01 – Turner – (56705) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Three point charges are located at the vertices oF an equilateral triangle. The charge at the top vertex oF the triangle is given in the fgure. The two charges q at the bottom vertices oF the triangle are equal. A Fourth charge 2 μ C is placed below the triangle on its symmetry- axis, and experiences a zero net Force From the other three charges, as shown in the fgure below. 7 . 4m 5m - 5 . 7 μ C qq 2 μ C ±ind q. Correct answer: 1 . 05392 μ C. Explanation: Let : Q 1 = - 5 . 7 μ C , Q 2 = q, Q 3 = Q 4 =2 μ C , a =7 . , and d = - . a d Q 1 Q 2 Q 3 2 μ C θ The Force on Q 4 is due to the Coulomb Forces From Q 1 , Q 2 , and Q 3 . Because Q 2 and Q 3 have equal charge, the x -components oF their Forces cancel out (by symmetry). Thus we only need to consider the y -components oF the Forces. Coulomb’s law tells us F i = k Q i Q 4 r 2 is the i th Force on Q 4 From the i th charge. The Forces From Q 2 and Q 3 are equal to each other, and opposite the direction oF the Force From Q 1 , since otherwise they could not cancel. Total Force on Q 4 is F Q 4 = k Q 1 Q 4 r 2 14 +2 k Q 2 Q 4 r 2 24 sin θ with r 14 the distance between Q 4 and Q 1 , r 24 = r 34 the distance between Q 4 and either Q 2 or Q 3 , and θ indicated in the sketch above. Remember that this Force F Q 4 will be set equal to zero since the problem tells us the Forces are in equilibrium. Because Q 1 , Q 2 , and Q 3 Form an equilateral triangle, oF sides oF length a , it can be seen that r 2 14 = ± 3 2 a + | d | ² 2 and r 2 24 = a 2 4 + d 2 . Also, sin θ = | d | r 24 = | d | ³ d 2 + a 2 4 . Our Force equation becomes 0= kQ 4 Q 1 ± 3 2 a + | d | ² 2

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howard (cah3459) – oldhomework 01 – Turner – (56705) 2 + 2 Q 2 ± a 2 4 + d 2 ² | d | ³ a 2 4 + d 2 . Rearranging, we get Q 2 = - Q 1 2 | d | ± a 2 4 + d 2 ² 3 / 2 ´ 3 2 a + | d | µ 2 = - ( - 5 . 7 μ C) 2 (5 m) × (7 . 4 m) 2 4 +( - 5 m) 2 · 3 / 2 ¸ 3 2 (7 . 4 m) + 5 m ¹ 2 = 1 . 05392 μ C . Note: Neither the sign nor the magnitude of the charge Q 4 (given in the problem as 2 μ C) enter into this equation. Actually, “the resultant force on Q 4 is zero” means that the resultant electric Feld is zero. Because the electric Fled is independent of the test charge Q 4 , the answer is also independent of the sign or magnitude of the charge Q 4 . 002 (part 1 of 2) 10.0 points Three point-charges ( - q ,+ q , and - q ) are placed at the vertices of an equilateral triangle (see Fgure below). a 60 - q - q + q ˆ ı ˆ The magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges is given by 1. ± ± F ± = 1 2 3 kq 2 a 2 2. ± ± F ± = 2 a 2 correct 3. ± ± F ± = 2 3 2 a 2 4. ± ± F ± = 3 2 2 a 2 5. ± ± F ± = 2 3 2 a 2 6. ± ± F ± = 3 2 2 a 2 7. ± ± F ± = 1 2 2 a 2 8. ± ± F ± = 1 2 2 a 2 9. ± ± F ± = 2 2 a 2 10. ± ± F ± = 3 2 a 2 Explanation: - q - q + q ˆ ı ˆ In this case, each of these forces has a mag- nitude F 21 = F 31 = 2 a
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oldhomework 01-solutions - howard(cah3459 oldhomework 01...

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