Final answer:
To prevent an electron with kinetic energy of 2.5 keV from being deflected in a 10 kV/m electric field, a magnetic field is required such that the magnetic force equals the electric force. For a proton to remain undeflected, it must travel at a speed where the forces balance, but due to its higher mass, the required magnetic field strength may differ.
Step-by-step explanation:
Magnitude and Direction of Uniform Magnetic Field for Electron
For an electron with kinetic energy of 2.5 keV to move horizontally in a downward-directed uniform electric field of magnitude 10 kV/m without deflection, we require a magnetic field that will exert a force equal in magnitude and opposite in direction to the electric force. The electric force (F_e) is qE, where q is the charge of the electron and E is the electric field strength. The magnetic force (F_b) on a moving charge is given by qvB, where v is the velocity of the charge and B is the magnetic field.
For F_e = F_b, we have qE = qvB. The charge cancels out, so E = vB. To find v, we use the kinetic energy (KE) of the electron: KE = (1/2)mv^2. We can solve for v and then B, ensuring the direction of B is perpendicular to both the electric field and the velocity of the electron (as per the right-hand rule), in this case into or out of the page.
Conditions for an Undeflected Proton
A proton can pass through the combination of fields undeflected if it experiences forces of equal magnitude but opposite direction from the electric and magnetic fields. This happens when E = vB for the proton as well, but since the proton has a different velocity for a given kinetic energy due to its greater mass compared to an electron, the magnetic field required to keep it undeflected could be different. However, if it moves at a speed such that the magnetic force equals the electric force acting on it, it will continue undisturbed.