Answer:
The angle is 89°.
Explanation:
Given that,
Electric field = 10 N/C
Electric flux = 0.01 N-m²/C
Area
We need to calculate the angle
Using formula of electric flux
![\phi=EA\cos\theta](https://tex.z-dn.net/?f=%5Cphi%3DEA%5Ccos%5Ctheta)
![\cos\theta=\dfrac{\phi}{EA}](https://tex.z-dn.net/?f=%5Ccos%5Ctheta%3D%5Cdfrac%7B%5Cphi%7D%7BEA%7D)
Where, E = electric field
= electric flux
A = area
Put the value into the formula
![\cos\theta=\dfrac{\dfrac{0.01}{2}}{10\times\pi\times(0.1)^2}](https://tex.z-dn.net/?f=%20%5Ccos%5Ctheta%3D%5Cdfrac%7B%5Cdfrac%7B0.01%7D%7B2%7D%7D%7B10%5Ctimes%5Cpi%5Ctimes%280.1%29%5E2%7D)
![\theta=\cos^{-1}(0.01592)](https://tex.z-dn.net/?f=%5Ctheta%3D%5Ccos%5E%7B-1%7D%280.01592%29)
![\theta=89.0^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D89.0%5E%7B%5Ccirc%7D)
Hence, The angle is 89°.
Answer:
We might get the benefit from this claim
Answer:
The magnitude of the acceleration of a proton at a distance of 1.5cm from the bead is 5.6 x10^13m/s².
The magnitude of the acceleration of a proton at a distance of 1.5cm from the bead is 9.8 x10^16m/s².
Explanation:
Newton's second law states that the total sum of the force acting on a particle in motion is equal to the mass of the particle times the acceleration due to the force. So the electric force between the bead and proton is equal to ma. That is,
Fe = kq1*q2/r² = m*a
The proton had a charge of +1.6x10^-19C and a mass of 1.67×10^-27kg
By substituting these values into the equivalent for a we have that the acceleration of the proton at a distance of 1.5cm form the bead is 5.6×10^13m/s²
The proton had a charge of -1.6x10^-19C and a mass of 9.10×10^-31kg
By substituting these values into the equivalent for a we have that the acceleration of the proton at a distance of 1.5cm form the bead is 9.8×10^16m/s²
A) ![x=\pm \frac{A}{2\sqrt{2}}](https://tex.z-dn.net/?f=x%3D%5Cpm%20%5Cfrac%7BA%7D%7B2%5Csqrt%7B2%7D%7D)
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:
![U=\frac{1}{3}K](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B3%7DK)
Using (2) we can rewrite this as
![U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B3%7D%28E-U%29%3D%5Cfrac%7B1%7D%7B3%7DE-%5Cfrac%7B1%7D%7B3%7DU%5C%5CU%3D%5Cfrac%7BE%7D%7B4%7D)
And using (1), we find
![U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7BE%7D%7B4%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7DkA%5E2%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B8%7DkA%5E2)
Substituting
into the last equation, we find the value of x:
![\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B8%7DkA%5E2%5C%5Cx%3D%5Cpm%20%5Cfrac%7BA%7D%7B2%5Csqrt%7B2%7D%7D)
B) ![x=\pm \frac{3}{\sqrt{10}}A](https://tex.z-dn.net/?f=x%3D%5Cpm%20%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7DA)
In this case, the kinetic energy is 1/10 of the total energy:
![K=\frac{1}{10}E](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B10%7DE)
Since we have
![K=E-U](https://tex.z-dn.net/?f=K%3DE-U)
we can write
![E-U=\frac{1}{10}E\\U=\frac{9}{10}E](https://tex.z-dn.net/?f=E-U%3D%5Cfrac%7B1%7D%7B10%7DE%5C%5CU%3D%5Cfrac%7B9%7D%7B10%7DE)
And so we find:
![\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%3D%20%5Cfrac%7B9%7D%7B10%7D%28%5Cfrac%7B1%7D%7B2%7DkA%5E2%29%3D%5Cfrac%7B9%7D%7B20%7DkA%5E2%5C%5Cx%5E2%20%3D%20%5Cfrac%7B9%7D%7B10%7DA%5E2%5C%5Cx%3D%5Cpm%20%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7DA)