Question

IP A charge of 21.0 μC is held fixed at the origin. Part A If a -7.00 μC charge with a mass of 3.50 g is released from rest at the position (0.925 m, 1.17 m), what is its speed when it is halfway to the origin? v = m/s Previous AnswersRequest Answer Incorrect; Try Again; 11 attempts remaining Part BPart complete Suppose the -7.00 μC charge is released from rest at the point x = 12(0.925m) and y = 12(1.17m). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A? Greater than the speed found in part A Lee than the speed found part in A Equal to the speed found in part A Previous Answers Correct Part C Find the speed of the charge for the situation described in part B. v = m/s

Answer 1

Part A.

Using energy conservation for two charge system:

q1 = 21.0*10^-6 C & q2 = -7.00*10^-6 C

KEi + PEi = KEf + PEf

KEi = Initial kinetic energy of system = 0, (since q1 is held fixed and q2 starts from rest)

KEf = final KE of system = 0 + (1/2)*m*V^2 (again due to q1 KE is zero, while V is speed of q2 and m is mass of q2)

PEi = k*q1*q2/ri

PEf = k*q1*q2/rf

now given that if ri = r, then rf = r/2, So

0 + k*q1*q2/r = (1/2)*m*V^2 + k*q1*q2/(r/2)

k*q1*q2/r = (1/2)*m*V^2 + 2*k*q1*q2/r

(1/2)*m*V^2 = -k*q1*q2/r

V = sqrt (-2*k*q1*q2/(m*r))

given that m = 3.50 gm = 3.50*10^-3 kg

r = distance between P (0.925, 1.17) m & O (0, 0) m = sqrt (0.925^2 + 1.17^2) = 1.49 m

So,

V = sqrt (-2*9*10^9*21.0*10^-6*(-7.00*10^-6)/(3.50*10^-3*1.49))

V = 22.5 m/s = Speed of q2 when it is halfway to the origin

Part B.

Now in this we know that: KEi = 0, KEf = (1/2)*m*V1^2

PEi = k*q1*q2/ri & PEf = k*q1*q2/rf, where if ri = r1, then rf = r1/2, then we will again get that:

V1 = sqrt (-2*k*q1*q2/(m*r1))

r1 = distance between P (0.925/2, 1.17/2) & O (0, 0) = sqrt ((0.925/2)^2 + (1.17/2)^2) = 0.745 m

Using given values:

V1 = sqrt (-2*9*10^9*21.0*10^-6*(-7.00*10^-6)/(3.50*10^-3*0.745))

V1 = 31.9 m/s = Speed of q2 when it is halfway to the origin

So we can see that: V1 > V

Which means Greater than speed found in part A.

Part C.

As calculated above in part B, speed of charge will be

V = 31.9 m/s = Speed of q2 when it is halfway to the origin

Let me know if you've any query.