t A charged particle (q-0.8mC) moves in a region where the only force acting on
ID: 1730539 • Letter: T
Question
t A charged particle (q-0.8mC) moves in a region where the only force acting on the particle is an electric force. If the particle is released from rest in point A and reaches point B with a kinetic energy of 4.8 J. What is the electric potential difference Vg-V,? 2. Points A-(2,3)m and B (5,7)m are in a region where the electric field is uniform and given by E-(4i+3j) N/C. What is the potential difference Vg-V, between these two points? 3. A particle of mass-6.7x1027kg and charge-3.2x10-1°C moves along the x axis in the positive direction with a speed of 4.8x105 m/s. It enters a region of uniform electric field parallel to its motion and comes to rest after moving 2.0 m into the field. What is the magnitude of the electric field? 4. Consider a 2 mC charge at the origin of coordinates, how much work is necessary to bring a 3mC charge from infinity to a point with coordinates (3,4). Is the work done by the external agent positive or negative? s. Consider a 2 mC charge at the origin of coordinates, what is the potential difference between A-(0,2) and B-(0.3)Explanation / Answer
1.Energy conservation,
PE_A + KE_A = PE_B + KE_B
(-0.8 mC) VA + 0 = (-0.8 mC)(VB) + 4.8
(0.8 x 10^-3 C) (VB - VA) = 4.8
VB - VA = 6000 Volt .....Ans
2. V = - E.r
E = 4i + 3j and r = (5-2)i + (7-3)j = 3i + 4j
V = - (4 x 3) - (3 x 4)= - 24 Volt
3. Applying work - energy theorem,
Work done = change in KE
q E d = m (vf^2 - vi^2)/2
(3.2 x 10^-19)(E) (2) = (6.7 x 10^-27)[ 0 - (4.8 x 10^5)^2]/2
E = 1206 N/C .....Ans
4. W = change in PE = k q1 q2 / d - 0
= (9 x 10^9)(2 x 10^-3)(3 x 10^-3) / sqrt(3^2 + 4^2)
= 10800 J
work done is positive,
5. V = k q / r
delta(V) = (9 x 10^9)(2 x 10^-3)[1/2 - 1/3]
= 3 x 10^6 Volt
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