Two 2.0-cm-diameter disks face each other, 1.0mm apart. They are charged to +/- 10nC.?

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What is the electric field strength between the disks?

A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?

3 Answers

  • The electric field between the plates is

    E = σ/εo… where σ = area charge distribution

    = 10X10^-9/(π*(0.010m)^2) = 3.183×10^-5C/m^2

    and εo = 8.854×10^-12C^2/N-m^2

    So E = 3.183×10^-5C/m^2/8.854×10^-12C^2/N-m^2 = 3.60×10^6N/C

    Using conservation of energy we solve the second part

    (K + U) b = (K + U)t…Here Ut – Ub = q*(Vt – Vb)…and Vt – Vb = E*d = 3.60×10^6V/m*1.0×10^-3m = 3595V

    Since Kt = 0 we have Kb = q*E*d = 1.60×10^-19C*3595 = 5.75×10^-16J

    So 1/2*m*v^2 = 5.75×10^-16J

    Therefore v = sqrt(2*5.75×10^-16J/m) = sqrt(2*5.75×10^-16J/1.67×10^-27kg) = 8.30×10^5m/s

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