Unformatted text preview: LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS TEST 4 Time: 50 minutes; Calculators & formulae provided at the end = only aid; Total = 20 points. 1) [5] Three identical charges with Q = +7.5µC are located at P1 = (−L, 0), P2 (L, 0) and P1 = (0, L) in the xy plane. What are the magnitude and direction of the force on an electron located at the origin O = (0, 0) of the plane? Begin with a ﬁgure which shows the location of the three charges, and the electron. 1 2) [5] A small plastic sphere carries an unknown charge Q. It is suspended by a string and placed in a uniform electric ﬁeld E = 100 ˆ N/C that is turned on gently at t = 0. One ﬁnds i that it ﬁnally makes an angle θ = 45◦ with respect to the vertical direction, and that it turned clockwise from the vertical as the ﬁeld was turned on. Determine the charge Q (magnitude and sign). Start with a ﬁgure depicting the initial and ﬁnal positions of the suspendend sphere and freebody diagrams for both positions. 2 3) [5] Two thin, largearea insulating planes have been charged uniformly and carry a surface density of σ = −10 nC/cm2 respectively. They are placed parallel to each other, oriented vertically, and separated by a distance of ∆x = 2.0 cm. Plate 1 has location X1 = −1.0 cm, and plate 2 is at X2 = 1.0 cm. A proton is placed at three possible locations: (a) in the middle (x = 0); (b) at x = 2.0 cm; (c) at x = −4.0 cm. Determine the electric force experienced by the proton at the three locations (magnitude and direction for each case). Start with a sideview drawing. There is no need to derive the electric ﬁeld from Gauss’ law, you can use the expressions provided on the formula sheet. 3 4) [5] A proton moves from a location where V = 125V to a place where V = −40V. (a) What is the change in the proton’s kinetic energy? (b) Now replace the proton by an electron. What is the change in KE in this case? 4 FORMULA SHEET v (tf ) = v (ti ) + ti a(t) dt s(tf ) = s(ti ) + ttif v (t) dt vf = vi + a∆t sf = si + vi ∆t + 1 a∆t2 vf2 = vi2 + 2a∆s 2
tf g = 9.8 m/s2 f (t) = t f (t) = a = 0 F (t) = f (t) dt = at + C F (t) = antiderivative = indeﬁnite integral area under the curve f (t) between limits t1 and t2 : F (t2 ) − F (t1 ) x2 + px + q = 0 factored by: x1,2 = − p ± p4 − q 2 uniform circular m. (t) = R(cos ω t ˆ + sin ω t ˆ); (t) = r i jv exp = exp; sin = cos; cos = − sin. m = Fnet ; a FG =
Gm1 m2 ; r2 d [f (g (x))] dx df dt df dt = 1 F (t) = f (t) dt = t2 2 +C 2 = d r dt df dg ; dg dx = ...; (t) = a d v dt = .... (f g ) = f g + f g
2 g= GME 2; RE a a in in ﬁn ﬁn ∆p1 + ∆p2 = 0 ; K1 + K2 = K1 + K2 for elastic collisions. CM = m1 1 +m2 2 a m1 +m2 2 ˆ = × F ; τz = rF sin(α) for , F in xy plane. I = i mi ri ; I αz = τz ; (k = rot. axis) τr r r τ Krot = I ω 2 ; Lz = I ωz ; d Lz = τz ; L = × p; d L = 2 dt dt fs ≤ µs n; fk = µk n; fr = µr n; µr << µk < µs . FH = −k ∆x = −k (x − x0 ). Fd ∼ − ; linear: Fd = dv ; quadratic: Fd = 0.5ρAv 2 ; A = cross sectional area v W = F ∆x = F (∆r) cos θ. W = area under Fx (x). P E H = k (∆x)2 ; P E g = mg ∆y . 2 avg ∆p = J = F (t)dt; ∆px = Jx = area under Fx (t) = Fx ∆t ; p = m ; K = m v 2 v 2 RE = 6370 km; G = 6.67 × 10−11 Nm2 ; ME = 6.0 × 1024 kg kg x(t) = A cos (ω t + φ); ω = 2π = 2π f ; vx (t) = ...; vmax = ... T me = 9.11 × 10−31 kg mp = 1.67 × 10−27 kg e = 1.60 × 10−19 C FC =
Kq1 q2 ˆ r r2 K= 1 4π0 FE = q E Eline = 2K λ r = 2K Q Lr Eplane = η  20 =
2 Q 2A0 Ecap = mv 2 + Uel (s) 2 = 2 mv0 + Uel (s0 ), 2 C C Q = C ∆VC farad = F = V C = 0dA 0 = 8.85 × 10−12 Nm2 − − − parallel C1 , C2 : Ceq = C1 + C2 series C1 , C2 : Ceq1 = C1 1 + C2 1 (U ≡ P E el ) Uel = qEx for E = −E ˆ i Vel = Uel /q = 9.0 × 109 Nm C2
Q , pos 0 A 2 → neg Ex = − dVel dx
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This note was uploaded on 09/28/2010 for the course MATH 1025 taught by Professor Tammie during the Fall '10 term at York University.
 Fall '10
 Tammie
 Linear Algebra, Algebra

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