(a) An electron starts at rest and accelerates through an electric field established by a set of parallel plates with a potential difference of 35 V. What is the speed of the electron the instant before it hits the negative plate?
(e = 1.6 × 10-19 C, melectron = 9.1 × 10-31 kg) (___/ 6)
(b) Instead of hitting the negative plate, the electron, travelling East, escapes the parallel plates through a small hole and enters a magnetic field of 0.75 T directed downward. What will be the magnetic force (magnitude and direction) on the charge?
(c) Once the electron has entered the magnetic field, it is in circular motion. What is the radius of the electron’s circular path?
(a) We can use the equation for the energy gain of an electron in an electric field:
ΔE = eΔV
where ΔE is the energy gained by the electron, e is the electron charge, and ΔV is the potential difference across the plates.
We know that the electron starts at rest, so its initial energy is zero. Therefore, its final energy is equal to the energy gained:
ΔE = 0.5mv^2
where m is the electron mass and v is its final speed.
Equating these two expressions for ΔE, we get:
eΔV = 0.5mv^2
Solving for v, we get:
v = sqrt(2eΔV/m)
Plugging in the given values, we get:
v = sqrt(2(1.6×10^-19 C)(35 V)/(9.1×10^-31 kg))
v ≈ 2.06×10^6 m/s
Therefore, the speed of the electron just before hitting the negative plate is approximately 2.06×10^6 m/s.
(b) The magnetic force on a charged particle moving in a magnetic field is given by the equation:
Fm = qvB
where Fm is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field strength.
Since the electron is negatively charged and moving east, its velocity is westward. Therefore, the magnetic force will be directed downwards.
Plugging in the given values, we get:
Fm = (1.6×10^-19 C)(2.06×10^6 m/s)(0.75 T)
Fm ≈ 2.46×10^-14 N
Therefore, the magnitude of the magnetic force on the electron is approximately 2.46×10^-14 N, and its direction is downward.
(c) The magnetic force on the electron provides the necessary centripetal force for it to move in a circular path. The centripetal force is given by:
Fc = mv^2/r
where Fc is the centripetal force and r is the radius of the circular path.
Equating this with the magnetic force, we get:
qvB = mv^2/r
Solving for r, we get:
r = mv/qB
Plugging in the given values, we get:
r = (9.1×10^-31 kg)(2.06×10^6 m/s)/(1.6×10^-19 C)(0.75 T)
r ≈ 7.28×10^-3 m
Therefore, the radius of the electron's circular path is approximately 7.28×10^-3 m.