Two particles are fixed on an x axis. Particle 1 of charge 49.9 C is located at

Question

Two particles are fixed on an x axis. Particle 1 of charge 49.9 C is located at x = -13.8 cm; particle 2 of charge Q is located at x = 4.81 cm. Particle 3 of charge magnitude 47.6 C is released from rest on the y axis at y = 13.8 cm. What is the value of Q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?

units should be in C

Chapter 21, Problem 012 tharge magnitude 47.5 acceleration of particle 3 is in the poeitive diretin of (a) he x axis and (b) e y axis? (a) Number (b) Numher Click if you would like to Show Work for this questio: pen Show Wark Units

Explanation / Answer

given, particle 1, Charge Q1 = 49.9 micro C, location x = -13.8 cm

given, particle 2, Charge Q, location x = 4.81 cm

given, particle 3, Charge Q3 = 47.6 micro C, location y = 13.8 cm

so in cartesian coordinates, electric field at location of Q3 is given by

E = kQ1(0.138i + 0.138j)/(0.138^2 + 0.138^2)^3/2 + kQ(-0.0481i + 0.0138j)/(0.138^2 + 0.0481^2)^3/2

a. if initial acceleration is along +ve x axis

then y component of E field is 0

kQ1( 0.138)/(0.138^2 + 0.138^2)^3/2 + kQ(0.0138)/(0.138^2 + 0.0481^2)^3/2 = 0

49.9*10^(-6)( 0.138)/(0.138^2 + 0.138^2)^3/2 = -Q(0.138)/(0.138^2 + 0.0481^2)^3/2

Q = - 20.953 micro C

b. if initial acceleration is along +ve y axis

then x component of E field is 0

kQ1(0.138 )/(0.138^2 + 0.138^2)^3/2 + kQ(-0.0481 )/(0.138^2 + 0.0481^2)^3/2 = 0

49.9*10^-6*(0.138 )/(0.138^2 + 0.138^2)^3/2 = Q(0.0481i )/(0.138^2 + 0.0481^2)^3/2

Q = 60.1147 micro C

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