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Material Type: Exam; Class: Physics 2 - Intermediate; Subject: Physics; University: Bellarmine University; Term: Forever 1989;
Typology: Exams
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P206 E1spr 2011 Feb 14, 2011
field at x = 1 m, y = 1 m, z = 1 m.
Soln: ФE = qin / o we see that the only charge that is inside the spherical surface of radius 1.5 m is q = -40 nC, hence ФE = ( - 40 x 10-9^ C ) / (8.85 x 10-12^ C^2 / N. m^2 ) = - 4.52 x 10^3 Nm^2 /C
P206 E1spr 2011 Feb 14, 2011 Q.3 [5] Two point charges are arranged such that q 1 = 6.00 nC is at (0.3 m, 0), q 2 = -3.00 nC is at (0, -0.1 m)
Soln: Use VP = k (q 1 /r + q 2 /r) = 8.99x10^9 (6x10-9/(.3) + (-3.0x10-9) / (0.1)) = -89.9 V
Q.4 [5] Points A and B lie 20 cm apart on a line extending radially
VA = kQ/r = 280 or kQ = 280 r ……(1) and VB = kQ /(0.2 + r) = 130 or kQ = 130 (0.2 + r) …..(2) From (1) and (2) 280 r = 130 (0.2 + r) get r = 0.173 m and Q = 5.39x10-9^ C Ans: r = 0.173 m and Q = 5.39x10-9^ C Q.5[5] How strong an electric field is needed to accelerate electrons in a TV tube from rest to one-tenth the speed
Soln: Method I: Δk = kk = kf – ki = ½ me * vf^2 – 0 = ½ * 9.1x10-31^ * (0.13x10^8 )^2 = 4.095x10-16^ ; ki , kinetic energy is zero initially. Potential energy change Δk = kU = - Δk = kk; and Δk = kU = q Δk = kV = - q* d* E ; d = distance travelled, E is the electric field Hence E = Δk = kk / (q* d) Calculate and get electric field , E = 5.1x10^4 V/m as the answer. Ans: E = 5.1x10^4 0.3 m q 1 -0.1m q X Y P Q A B r