Question

3. + -/2 points SerCP10 15P005 My Notes Ask Your Te The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and neutrons) (a) what is the force between the two alpha particles when they are 5.60 × 10-15 m apart? (b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s

need all these 3 questions ...just need final answers only

Answer 1

Solution:

(3) Solution:

According to the problem, alpha particles are having a charge of 2 protons.

Charge of alpha particles = 2 X 1.6 X 10^-19 C = 3.2 X 10^-19 C

(a) Solution:

Force between two charged particles can be calculated as follows

$F=\frac{1}{4\pi \varepsilon _{o}}\frac{3.2\times 10^{-19}\times 3.2\times 10^{-19}}{5.6^{2}\times 10^{-30}}$

$F=29.39\; \mathrm{N}$

(b) Solution:

Mass of alpha particle, m = 4.0026 u = 4.0026 X 1.67 X10^-27 = 6.684 X 10^-27 kg

Force, F = 29.39 N

Acceleration can be calculated as follows

$F=ma$

$a=\frac{F}{m}=\frac{29.39}{6.684\times 10^{-27}}$

$a=4.397\times 10^{27}\; \mathrm{m/s^{2}}$

(7) Solution:

Charge of the sphere, q = 67 X 10^-6 C

Mass of the sphere, m = 5.2 g = 5.2 X 10^-3 kg

Angle made by Electric field with horzontal $\theta =35^{\circ}$

(a) Solution:

Free body diagram of the given problem is shown in below

(b) Solution:

From above free body diagram we can write

$E\: \mathrm{sin}\: \theta =mg$

$E=\frac{mg}{\mathrm{sin}\: \theta }$

$E=\frac{5.2\times 10^{-3}\times 9.8}{\mathrm{sin}\: 35 }$

$E=8.85\times 10^{-4}\; \mathrm{N/C}$

(c) Solution:

Tension can be calculated as follows

$T=E\: \mathrm{cos}\: \theta$

$T=8.85\times 10^{-4}\times \mathrm{cos}\:35$

$T=7.249\times 10^{-4}\; \mathrm{N}$

(10) Solution:

(a) Solution:

Electric field due to 3.00 nC at point 'P' is given by

$E_{1}=\frac{1}{4\pi \varepsilon _{o}}\frac{q}{r^{2}}$

$E_{1}=\frac{1}{4\pi \varepsilon _{o}}\frac{3\times 10^{-9}}{4.36^{2}\times 10^{-4}}$

Simplifying this,

$E_{1}=1.420\times 10^{-5}\; \mathrm{N/C}$

Same electric field will be exerted by another positve charge 3.0 nC.

I.e. $E_{3}=1.420\times 10^{-5}\; \mathrm{N/C}$

These twe fields will be acted towards the point 'P'

Now, we need to calculate the field due to charge -2.0 nC at point P.

$E_{2}=\frac{1}{4\pi \varepsilon _{o}}\frac{q}{r^{2}}$

$E_{2}=\frac{1}{4\pi \varepsilon _{o}}\frac{(-2)\times 10^{-9}}{4.36^{2}\times 10^{-4}}$

$E_{2}=-0.947\times 10^{-5}\; \mathrm{N/C}$

Negative sign indicates that the field will acts away from the point 'P'

resultant field at point 'P' can be calculated by resolving the field in to different components as shown in below figure.

+3.00 nC 2 -2-00 nC 30° 30° E, 2- 3 3 +3.00 nC

From this we can get the resultant field is as follows

$E_{resu}=E_{1}\: \mathrm{cos}\: \theta+E_{3}\: \mathrm{cos}\: \theta +E_{2}$

Since E₁ = E₃

$E_{resu}=2E\: \mathrm{cos}\: \theta +E_{2}$

$E_{resu}=(2\times 1.42\times 10^{-5}\times \: \mathrm{cos}\: 30) -(0.947\times 10^{-5})$

$E_{resu}=1.512\times 10^{-5}\; \mathrm{N/C}$

Positive sign indicates the the field is acting towards right.

(b) Solution:

Charge place at point 'P' is Q = -4.60 nC

We know the formula,

$F=-1.512\times 10^{-5}\times 4.60\times 10^{-9}$

$F=-6.969\times 10^{-14}\; \mathrm{N}$

Here, negative sign indicates that the force is acting to the left.