Deals with the concepts that are important in this problem. As illustrated in th

Question

Deals with the concepts that are important in this problem. As illustrated in the figure, a negatively charged particle is released from rest at point B and accelerates until it reaches point A. The mass and charge of the particle are 2.75 x 10*4 kg and -5.27 x 10-5 C, respectively. Only the gravitational force and the electrostatic force act on the particle, which moves on a horizontal straight line without rotating. The electric potential at A is 36.0 V greater than that at 8; in other words, VA - VB = 36.0 V. What is the translational speed of the particle at point A?

Explanation / Answer

Apply, conservation of energy

kA + UA = kB + UB

kA + UA = 0 + UB

kA = UB - UA

KA = q*(VB - VA)

= -q*(VA - VB)

= -(-5.27*10^-5)*36

= 0.001897 J

now use, kA = 0.5*m*vA^2

==> vA = sqrt(2*kA/m)

= sqrt(2*0.001897/(2.75*10^-4))

= 3.71 m/s <<<<<<<<<<<<<<<<<<<<<<<<<<<--------------Answer

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