Physics, 14.09.2019 01:10 KHaire2077

Two 2.5-cm-diameter disks face each other, 1.5 mm apart. they are charged to ± 14 nc. part a.
what is the electric field strength between the disks?
express your answer in newtons per coulomb.
part b.
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?
express your answer in meters per second.

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Answer from: diegorivas7992

Part A: The electric field strength between the discs = $\rm 1.582\times 10^3\ N/C.$

Part B: The launch speed of the proton = $2.132\times 10^4\ \rm m/s.$

Explanation:

Constant used:

$\epsilon_o$ = electrical permittivity of free space = $\rm 8.85\times 10^{-12}\ C^2N^{-1}m^{-2}.$

Assumptions and Given values:

Diameter of each disc, $\rm D= 2.5\ cm = 2.5\times 10^{-2}\ m.$ Distance between the discs, $\rm d = 1.5\ mm = 1.5\times 10^{-3}\ m.$ Charge on the positive plate, $\rm q_1 = +14\ nC = +14\times 10^{-9}\ C.$ Charge on the negative plate, $\rm q_2 = -14\ nC = -14\times 10^{-9}\ C.$

It is clear that the distance between the discs is much less than the diameter of each disc, therefore, the electric field in the region between the discs is nearly uniform.

According to the Gauss' law of electrostatics, the strength of the electric field due to charged disc at a point r distance away is given by

$\rm E=\dfrac{|q|}{2\epsilon_o}$

where, q is the charge on the disc and this expression is only valid if

r << (Diameter of the disc).

Assuming, the disc on left contains positive charge and the disc on right contains negative charge.

Part A:

The strength of electric field between the discs due to the left disc is given by

$\rm E_1 = \dfrac{|q_1|}{2\epsilon_o}.$

The left disc is positively charged therefore the direction of this field is towards right.

Similarly, the strength of electric field between the discs due to the right disc is given by

$\rm E_2 = \dfrac{|q_2|}{2\epsilon_o}.$

The right disc is negatively charged therefore the direction of this field is also towards right.

Since, the electric fields due to both the discs are towards right therefore the strength of the net electric field between the discs due to both the discs is given by

$\rm E = E_1+E_2\\=\dfrac{|q_1|}{2\epsilon_o}+\dfrac{|q_2|}{2\epsilon_o}\\=\dfrac{|q_1|+|q_2|}{2\epsilon_o}\\=\dfrac{14\times 10^{-9}+14\times 10^{-9}}{2\times 8.85\times 10^{-12}}\\=1.582\times 10^3\ N/C.$

Part B:

We know,

Mass of the proton, $\rm m=1.67\times 10^{-27}\ kg.$ charge on the proton, $\rm q=+1.6\times 10^{-19}\ C.$

According to Coulomb's law, the electrostatic force on the proton, between the discs, is given by

$\rm F=qE=1.6\times 10^{-19}\times 1.582\times 10^3=2.53\times 10^{-16}\ N.$

According to Newton's second law of motion,

$\rm F=ma\\\Rightarrow a = \dfrac Fm=\dfrac{2.53\times 10^{-16}}{1.67\times 10^{-27}}=1.515\times 10^{11}\ m/s^2.$

It is the acceleration acquired by the proton between the discs.

The direction of the acceleration is from positive plate to negative plate.

When the proton moves from negative plate to positive plate its effective acceleration would be

$\rm a_{eff} = -a = -1.515\times 10^{11}\ m/s^2.$

Let the launch speed of the proton from the negative disc be u.

Since, the proton barely reach the positive plate therefore its speed when it touches the positive plate is 0, i.e., v = 0 m/s.

Using the equation,

$\rm v^2-u^2=2a_{eff}d\\0^2-u^2=2\times (-1.515\times 10^{11})\times 1.5\times 10^{-3}\\u^2=4.545\times 10^8\\u=2.132\times 10^4\ m/s.$

It is the required launch speed.

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Two 2.5-cm-diameter disks face each other, 1.5 mm apart. they are charged to ± 14 nc. part a.
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