Final No Date 2

Prep101comsolutions 6 prep101 phys1026 april exam

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Unformatted text preview: 2 =⎜2 ⎜d ⎝ 2 ⎞ ⎛ kQ 2 ⎟ +⎜ 2 ⎟ ⎜d ⎠⎝ 2 ⎞ ⎛ kQ 2 ⎟ = 2⎜ 2 ⎟ ⎜d ⎠ ⎝ 2 ⎞ kQ 2 ⎟=2 2 N ⎟ d ⎠ Force = Fnet = 2 FBonA kQ 2 N [x 45°y] d2 (b) This is a conceptual question. If you wanted to, you could do some really tricky math to obtain an equation for the coordinates of some general point in space where the field is zero, but you won’t be able to solve it. The electric field is only zero in a few Solutions will be posted at www.prep101.com/solutions 6 ©Prep101 Phys1026 April Exam Solutions special cases: at infinity; in the absence of all charges; or in certain very symmetrical cases. Our particular case is none of these, so you should answer, NO. Something that could tip you off to the answer is the fact that the question says “If so, how many points are there, and where are they located? (Do not find the coordinates)”. Where are the points, but don’t tell us where they are…that’s a contradiction. Points where field is zero = NONE Answer 2.5 kq1 q 2 ∴ FE α q1, FE α q2, FE α 1/r2 2 r a) q1 → 2q1; q2 → 2q2; r=r 4kq1 q 2 → 4(3.0 x 10-6N) = 1.2 x 10-5N ∴ FE → r2 b) uncharged sphere will ‘accept’ half of the charge from q1 kq q ∴ FE → ½ 12 2 → ½(3.0 x 10-6N) = 1.5 x 10-6N r c) r → 3r kq1 q 2 ∴ FE → → (3.0 x 10-6N)/9 = 3.3 x 10-7N (3r ) 2 Answer 2.6 For equilibrium, F13 = F23 kq1 q3 kq 2 2q3 = --- k, q3 cancel out 2 x (2 − x) 2 1.6x10-5/x2 = 6.4x10-5/(2-x)2 1/ x2 = 4/(2-x)2 (2-x)2 = 4x2 4 – 4x – x2 = 4x2 ---use quadratic formula to solve for roots 3x2 + 4x –4 = 0 FE = − b ± b 2 − 4ac 2a − 4 ± 16 − 4(3)(−4) x= 2(3) x = -2.0 or x = 0.67 ∴ the 3rd sphere must be 0.67m from the 1st sphere. x= Solutions will be posted at www.prep101.com/solutions 7 ©Prep101 Phys1026 April Exam Solutions ELECTRIC FIELDS PRACTICE QUESTIONS Question 1: Solution: In order for the net electric field to be zero, we need the fields due to the positive and negative charges to be of equal magnitude, but directed in opposite directions. The field due to the positive charge is going to point away from the positive charge; the field due to the negative charge is going to point towards the negative charge. Based on this knowledge alone, we know that our field point P cannot be between the charges. In this region both the electric fields are going to be pointing to the left. Based on the magnitude of the charges, we also know that the field point must be closer to the smaller charge, since the field due to the larger charge is going to be larger in magnitude. As indicated in the illustration, this means that the field point must be located to the left of the charges. Now, we can solve for the location of P by requiring that E + = E − : (6 .0 μ C ) 1 (− 2 .5 μ C ) 1 = 2 4πε o 4πε o (1m + d ) 2 d Now we just need to solve for d: 2 .5 (1m + d ) 2 = 6 .0 d 2 1m + d = ± 6 .0 = ± 1 .55 d 2 .5 and, isolating d we find d= 1m = 1.8m + 1.55 − 1 and d= 1m = −0.39m − 1.55 − 1 In thi...
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