Question

A very long nonconducting cylinder of diameter 10.0 cm carries charge distributed uniformly over its surface. Each meter of length carries +5.50 µC of charge. A protonis released from rest just outside the surface. How far will it be from the surface of the cylinder when its speed has reached 2550 km/s? The mass of the proton is1.67 × 10-27 kg and e = 1.60 × 10-19 C.

Answer 1

Given that diameter of the cyllinder is d = 10.0 cmRadius of the cyllinder is R = 5.0cmcharge density is λ = 5.50 *10^-6 C / mspeed of the proton is v = 2550 km /s= 2.55*10^6 m /sElectric potential at a distance x isV = q λ / 2 π ε_0 ln ( R / x )The potential energy of the proton is P.E = q λ / 2 π ε_0 ln ( R / x )P.E = K.Eq λ / 2 π ε_0 ln ( R / x ) = 1/ 2 m v^21.6*10^-19 C * 5.50*10^-6 C /m / 2 π *8.85*10^-12 ln ( 5.0*10^-2 m / x)= 1/2 * 1.67*10^-27 kg * ( 2.55*10^6 m/s)^2ln ( 5.0*10^-2 m / x) = 0.342( 5.0*10^-2 m / x) = 1.40x = 3.57 *10^-2 m= 0.0357 cm

Answer 2

The diameter of the cylinder is D = 10 cmThe radius of the cylinder is R = 5 cmThe linear charge density λ = 5.50 μC/mThe electric potential at a distancer from the axis of infinite long cylinder of radiu R carrying uniformly distributed charge density λ is givenbyV = λ/2πε₀ ln(R/r)Where ε₀ = 8.85*10^-12 C²/Nm² is permitivity of free spaceThe potential energy of the proton at this point will beU = eλ/2πε₀ ln(R/r)Where e = 1.6*10^-19 C is the charge of protonThe proton has a speed v = 2550 km/sThe change kinetic energy of the proton is ΔKE = 0.5mv²Where m = 1.67*10^-27 kgis the mass of the protonTherefore, according to the conservation of energy,U = ΔKEeλ/2πε₀ ln(R/r) = 0.5mv²(1.6*10^-19 C)(5.50*10^-6 C/m)/(2π)(8.85*10^-12 C²/Nm²)ln(R/r) = (0.5)(1.67*10^-27kg)(2.55*10⁶ m/s)²ln(R/r) = 0.3429R/r = 1.409r = 5 cm/1.409r = 3.548 cm

Answer 3

according to Gauss' law
Q/epsilon0=E*L*2pi*r.
where Q=5.5e-6*L.
we have
E=5.5e-6/2pi*epsilon0*r=2k*5.5e-6/r
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force exerted on the proton.
F=2k*5.5e-6*1.6e-19/r=1.58e-14/r(N).
kinetic energy of the proton.
K=mv^2/2=1.67e-27*2550e3^2/2=5.43e-15(J).
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we have that
F*dr=dA
1.58e-14*dr/r=dA
1.58e-14*ln(r1/r0)=5.43e-15
where r0=5cm.
so r1=7.1(cm)