v =-5.00 × 10\"ms-Ty Question 4: i) 2.0cm 2.0cm +10V 2.0cm The figure above show
ID: 1773605 • Letter: V
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v =-5.00 × 10"ms-Ty Question 4: i) 2.0cm 2.0cm +10V 2.0cm The figure above shows another uniform electric field in which an electron at point A has a velocity of 5.00 x 106 ms1 in the -y direction. Calculate: (a) the electric field, (b) the change in electric potential in moving from point A to B, (c) the change in the electric potential energy of the electron in moving from point A to B, (d) the kinetic energy of the electron at point B [hint: work out the kinetic energy at point A and use the law of conservation of energy and your answer to part (c)], and (e) the speed of the electron at point B. (mass of an electron 9.11 x 103 kg) Question 4: ii) 4.00 mm 4.00 mm +5.00 uC 5.00 uC x-axis 0 q2 source source charge origin charge A certain dipole consists of two source charges + 5.00 and-5.00 uC placed equidistant from the origin, along the x-axis as shown above. Find (a) the electric field at the origin, (b) the electric force on a test charge of +8e (e 1.6 x 10.19 C) placed at the origin, and (c) the electric potential at the origin If the-5.00 uC source charge is now changed to a 5.00 uC source charge, what is (d) the new electric field at the origin and (e) the new electric potential at the origin?Explanation / Answer
4. vA = -5 x 106 m/s y^
a) E = -dV/dr = -(25 - 20)/0.02 = -250 V/m y^
b) VB - VA = 10 - 25 = -15 V
c) change in potential energy of electron P = q(VB - VA) = -1.602 x 10-19 x (-15) = 24.03 x 10-19 J
d) KB + P = KA
=> KB = 0.5 x 9.11 x 10-31 x (5 x 106)2 - 24.03 x 10-19 = 89.845 x 10-19 J
e) vB = sqrt(2KB/m) = sqrt(2 x 89.845 x 10-19/9.11 x 10-31) = 4.441 x 106 m/s
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