Physics, 31.03.2020 02:20 2019dawnmcmillan

6. A proton is released from rest 1.0 cm off a center of a uniformly charged disk of a 2.0 m radius carrying a +180 nC charge. What is the speed of the proton when it is 5 cm from the disk?

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Answer from: yadirachavez2002

the speed of the proton is v = 78.1 km/s

Explanation:

Generally Electric potential energy resulting from a proton can be mathematically represented as

$V(z) = \frac{Q(\sqrt{R^2 + z^2} - z )}{2 \pi \epsilon_o R^2}$

Where

R is the radius of the disc which is given as R = 2m

q is the charge on the proton with a value of $q = 1.602 *10^{-19}C$

the mass of this proton has a value of $m=1.673*10^{-27}kg$

z is the distance of the center of the disk from the question

$z_1 =1.0cm = \frac{1}{100} = 0.01m$

$z_2 = 5cm = \frac{5}{100} = 0.05m$

$\epsilon_0 = 8.85*10^{-12}F/m$

According the law of conservation of energy

The change in kinetic energy of the proton + change in electrostatic = 0

potential energy

The change in kinetic energy is $\Delta KE = \frac{1}{2} m (v^2 - u^2)$

Since u = 0 the equation becomes

$\Delta KE = \frac{1}{2} m (v^2)$

change in electrostatic potential energy is = q (V(z_1) - V(z_2))

Substituting this into the equation we have

$\frac{1}{2} mv^2 +q(V(z_1)- V(z_2)) = 0$

$\frac{1}{2} mv^2 = -q(V(z_1)- V(z_2))$

Recalling that $V(z) = \frac{Q(\sqrt{R^2 + z^2} - z )}{2 \pi \epsilon_o R^2}$ we have

$\frac{1}{2} mv^2 = -q(\frac{Q(\sqrt{R^2 +z_2^2}-z_2 )}{2 \pi \epsilon_0 R^2} - \frac{Q(\sqrt{R^2 +z_1^2}-z_1 )}{2 \pi \epsilon_0 R^2} )$

$\frac{1}{2} mv^2 = q(\frac{Q(\sqrt{R^2 +z_1^2}-z_1 )}{2 \pi \epsilon_0 R^2} - \frac{Q(\sqrt{R^2 +z_2^2}-z_2 )}{2 \pi \epsilon_0 R^2} )$

Substituting values $\frac{1}{2} mv^2 = 1.602*10^{-19}(\frac{180*10^{-9}(\sqrt{2^2 +0.01^2}-0.011 )}{2 \pi 8.85*10^{-12}* 2^2} - \frac{180*10^{-9}(\sqrt{2^2 +0.05^2}-0.05 )}{2 \pi *8.85*10^{-12} *2^2} )$

Making v the subject

$v=\sqrt{\frac{1.602*10^{-19}(\frac{180*10^{-9}(\sqrt{2^2 +0.01^2}-0.011 )}{2 \pi 8.85*10^{-12}* 2^2} - \frac{180*10^{-9}(\sqrt{2^2 +0.05^2}-0.05 )}{2 \pi *8.85*10^{-12} *2^2} )}{0.5 *1.673*10^{-27}}}$

v = 78.1 km/s

Answer from: Quest

Iread this book last year so take this with a grain of salt.gene jumps from the tree to compete with finny due to the fact that gene is always trying to do better than finny. he is put under the grasp of peer pressure due to finny almost forcing him to do this. ever since he first jumped finny kept pressuring him to jump since he thought gene enjoyed jumping.rather than uses this answer you should probably just reread chapters 3-4

Answer from: Quest

B) as the temperature increases, the kinetic energy of the particles increases.|-/

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