Two 4.0-cm-diameter disks face each other, 1.5 mm apart. They are charged to ± 1

Question

Two 4.0-cm-diameter disks face each other, 1.5 mm apart. They are charged to ± 15 nC .

a.) What is the electric field strength between the disks?

b.) What is the magntiude of the electric force acting on a proton due to the electric field between the discs?

c.) 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? Assume the weight of the proton is negligible.

This is not the correct answer

this is also not the correct answer

b)force on proton = q*E = 1.6*10^-19*2.69*10^6 = 4.304*10^-13

c)energy = potential energy due to electric field = - q( E.dr) = -1.6*10^-19*2.60*10^6*(-1.5*10^-3) =6.456*10^-16J

Explanation / Answer

The electric field between the plates is
E = /o... where = area charge distribution = 15X10^-9/(*(0.020m)^2) = 1.194 x 10^-5 C/m^2
and o = 8.854x10^-12 C^2/N-m^2

So E = 1.194 x10^-5/8.854x10^-12 = 1.35 x 10^6 N/C

b) Force = F = qE = 1.6*10^-19*1.35 x 10^6 = 2.16*10^-13 N

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 = 1.35x10^6V/m*1.5x10^-3m = 2025 V
Since Kt = 0 we have Kb = q*E*d = 1.60x10^-19C*2025 = 3.24x10^-16J
So 1/2*m*v^2 = 3.24 x10^-16 J
Therefore v = sqrt(2*3.24x10^-16J/m) = sqrt(2*3.24x10^-16J/1.67x10^-27kg) = 6.23x10^5m/s

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