P23_043 - 43. (a) The electric field is upward in the...

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Unformatted text preview: 43. (a) The electric field is upward in the diagram and the charge is negative, so the force of the field on it is downward. The magnitude of the acceleration is a = eE/m, where E is the magnitude of the field and m is the mass of the electron. Its numerical value is a= (1.60 × 10−19 C)(2.00 × 103 N/C) 2 = 3.51 × 1014 m/s . 9.11 × 10−31 kg We put the origin of a coordinate system at the initial position of the electron. We take the x axis to be horizontal and positive to the right; take the y axis to be vertical and positive toward the top of the page. The kinematic equations are 1 y = v0 t sin θ − at2 , 2 x = v0 t cos θ , and vy = v0 sin θ − at . First, we find the greatest y coordinate attained by the electron. If it is less than d, the electron does not hit the upper plate. If it is greater than d, it will hit the upper plate if the corresponding x coordinate is less than L. The greatest y coordinate occurs when vy = 0. This means v0 sin θ −at = 0 or t = (v0 /a) sin θ and ymax = = 2 2 2 v0 sin2 θ 1 v0 sin2 θ 1 v0 sin2 θ = −a a 2 a2 2 a 6 2 (6.00 × 10 m/s) sin2 45◦ = 2.56 × 10−2 m . 14 m/s2 ) 2(3.51 × 10 Since this is greater than d = 2.00 cm, the electron might hit the upper plate. (b) Now, we find the x coordinate of the position of the electron when y = d. Since v0 sin θ = (6.00 × 106 m/s) sin 45◦ = 4.24 × 106 m/s and 2 2 2ad = 2(3.51 × 1014 m/s )(0.0200 m) = 1.40 × 1013 m2 /s the solution to d = v0 t sin θ − 1 at2 is 2 t= v0 sin θ − 2 v0 sin2 θ − 2ad a 4.24 × 10 m/s − 6 = = 2 (4.24 × 106 m/s)2 − 1.40 × 1013 m2 /s 2 3.51 × 1014 m/s 6.43 × 10−9 s . The negative root was used because we want the earliest time for which y = d. The x coordinate is x = v0 t cos θ = (6.00 × 106 m/s)(6.43 × 10−9 s) cos 45◦ = 2.72 × 10−2 m . This is less than L so the electron hits the upper plate at x = 2.72 cm. ...
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