(a) [tex]4.26\cdot 10^5 V/m[/tex]
In a velocity selector, the speed of the beam is related to the magnitude of the electric field and of the magnetic field by the formula:
[tex]v=\frac{E}{B}[/tex]
where
E is the magnitude of the electric field
B is the magnitude of the magnetic field
In this problem, we have
[tex]B=0.115 T[/tex] (magnetic field)
[tex]v=3.7\cdot 10^6 m/s[/tex] (speed of the particles)
Solving the equation for E, we find the electric field:
[tex]E=vB=(3.7\cdot 10^6 m/s)(0.115 T)=4.26\cdot 10^5 V/m[/tex]
(b) 3.2 kV
The relationship between electric field and potential difference between the two plates is:
[tex]V=Ed[/tex]
where, in this problem:
[tex]E=4.26\cdot 10^5 V/m[/tex] is the magnitude of the electric field
[tex]d=0.75 cm=0.0075 m[/tex] is the separation between the plates
Substituting into the equation, we find the potential difference:
[tex]V=(4.26\cdot 10^5 V/m)(0.0075 m)=3195 V=3.2 kV[/tex]
