1) In the figure above, Qa = +2.345 C and Qc= -2.345 C. Where Qa is is point A,
ID: 1586834 • Letter: 1
Question
1) In the figure above, Qa = +2.345 C and Qc= -2.345 C. Where Qa is is point A, where Qb is point B, etc.
A.What is the potential at point B due to Qa and Qc?
B. If Qb= +3.454 C, how much work does it take to move charge Qb from infinity to point B?
C.What is the electric potential energy of Qb when it is at point B?
D. If Qb has a mass of 3.1415 g, and its potential energy at point B is fully converted to kinetic energy, what is its speed?
E.What is the magnitude and direction of the electric field at point D due to Qa and Qc?
F.Suppose that a particle enters point D with a velocity that points along the +x axis, what are the allowed range of velocity directions with which it can leave point D?
x0.7457 m x,= 3.0481 m I D yd 1.7524 m IT77 yc= 1.0640 m -. eo,- ,-0.6799 m -Explanation / Answer
A.) distance between points A and B be rab and between B and C be rbc
rab = ( ( xb - xa )2 + (yb - ya )2 ) 1/2 by pythogoras theoram
rab = ( ( 0.7457 - 0 )2 + (0.6799 - 0 )2 ) 1/2 = 1.0091
similarly rbc = ( ( xb - xc )2 + (yb - yc )2 ) 1/2
rbc = ( ( 0.7457 - 3.0481 )2 + (0.6799 - 1.0640 )2 ) 1/2 = 2.33422
potential at B due to A and C is (1/4o) x (Qa / rab + Qc / rbc ) = 9 x 109 x ( 2.345 / 1.0091 - 2.345 / 2.3342)
= 1.1873 x 1010 V
b.) The work done to bring charge B to point B = Potential at point B x Charge of B
W = 1.1873 x 1010 x 3.454 = 4.1 x 1010 Joules
c.) Electric potential energy of Qb is nothing but the amount if work required to bring it to position B from infinity
i.e PE of Qb = W = 4.1 x 1010 Joules
d.) PE = KE = 4.1 x 1010 Joules
We know that KE = 0.5mv2 , where m = mass and v is the velocity
m= 3.1415 g = 3.1415 x 10-3 kg
=> 4.1 x 10 10 = 0.5 x 3.1415 x 10-3 x v2
=> v2 = 2.610820442 x 1013
v = 5.109618 x 106 m /s
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