Consider a horizontal parallel plate capactor where the size of the plates is mu

Question

Consider a horizontal parallel plate capactor where the size of the plates is much larger than the spacing between the plates. Between the plates is placed a small charged bead with a mass 2.1x10-12kg and a charge equal to that of 90 electrons. The separation between the plates is 9.8 cm and the charge per area on the plates is +/- 4.1x10-6C/m2. The top plate is positively charged and the bottom plate is negatively charged.

If the bead is released from rest halfway between the two plates, what is the speed of the bead when it hits one of the plates?

Give you answer in m/s three digits and do not include the units in your answer.

Explanation / Answer

Charge on the particle = 90 * 1.6 * 10^-19 = 1.44 * 10^-17 C

Electric Field due to the 2 plates = Q/epsilon , where Q is the charge density on the plate = 4.1 * 10^-6/(8.85 * 10^-12) = 4.63 * 10^5 N/C

Force = qE = 1.44 * 10^-17 * 4.63 * 10^5 = 6.67 * 10^-12 N

Acceleration of the body = (mg - qE)/m = (2.1 * 10^-12 * 9.8 - 6.67 * 10^-12)/(2.1 * 10^-12) = 6.623 m/s^2

Distance to be traveled = 4.9 cm = 4.9 * 10^-2 m

v^2 = u^2 + 2as

v^2 = 2 * 6.623 * 4.9 * 10^-2

v = 0.8 m/s

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