An electron starts from rest near the negative vertical plate of a set of parallel plates and accelerates towards the positive plate through 150 V. After crossing this distance the electron passes through a hole. The electron then moves through a horizontal set of parallel plates where the top plate is negative. These horizontal plates are separated by 2.0 cm and have a potential difference of 240 V. The electron passes through these plates undeflected due to a magnetic field perpendicular to the page. The electron then passes out of the plates where it is acted upon by the same uniform magnetic field. Find the magnitude and direction of the uniform magnetic field and the radius of the circular path of the electron. Draw the complete path of the electron.

-25 1.58 x 10 A phosphate molecule PO of mass m = difference |AV| = 4.35 V and enters the mass spectrometer with magnetic field of magnitude |B| = 0.090 T. The particle then follows a semi-circular path shown in the picture before kg is accelerated across the potential reaching the detector located to the left of its entry point. В d Δν a) Sketch the described situation (you can start with the figure shown in the problem) CLEARLY indicating the direction of the magnetic field inside the mass spectrometer b) Determine the speed of the particle as it enters the region of magnetic field. c) Determine the distance between the entry point of the particle and the detector.

An electron with a kinetic energy of 1.04 x 104 eV is initially moving south in a uniform magnetic field of magnitude 488 G, directed east. Describe in detail the electron's trajectory, including its shape and orientation. the radius of the path   direction of initial force        shape of trajectory

An electron is accelerated from rest through a potential difference V after which it enters a region having a magnetic field of magnitude B and direction perpendicular to the electron's velocity. The radius of the electron orbit in this magnetic field is given by: meV 2BN e 2m, V Be 2m. VB O 1 2m.V B e O 1 meV BV 2e O there is not enough information to determine 2m.V BV O none of these

F=0.8975r^-2.00 And re-writing it as:         (4)F=0.8975/r^-2.00     We can now compare it with Coulomb’s equation:         (1)F=K|q1q2|/r^2     So, comparing these equations (4) and (1) we realize that: k=|q1q2|=0.8975   Plugging in values for charges: q1 = -10µC, and q2 = 10µC, isolating  k (Coulomb’s constant), and recalling that 1µC = 10-6 C, we get: k|-10µC*10µC|=0.8975 k=0.8934/|-10µC*10µC|= 0.8934/10*10^-6C*10*10^-6C (3)   Use this Example calculation above, and the trendline equation (3) coefficients that you have obtained on your graph, to find k.  Show your work below. Be careful with powers of ten.      k=  0.8975/|-10µC*10µC|= 0.8975/10*10^-6C*10*10^-6C =     (5)       Now, important part in reporting the answer is properly showing its units. To derive the SI units for Coulomb’s constant k, let’s look again at the equation (1), and re-write it as:       k=F*r^2/|q1q2| (6)     Recalling the units for all parameters in this equation: Force, F – Newtons, [N] Distance, r –…