Answer:
V = 2.14×10⁵ V.
W = 3.424×10⁻¹⁴ J.
Explanation:
Electric Potential: This can be defined as the work done in bringing a unit positive charge from infinity to that point, against the action of a field.
The S.I unit is V.
The expression containing electric potential, distance and electric field is given as,
V = E×r .............. Equation 1
Where V = Electric potential difference across the length of the accelerator's section, E = Electric Field, r = Length of the section.
Given: E = 6.03×10⁵ N/C, r = 35.5 cm = 0.355 m.
Substitute into equation 1
V = 6.03×10⁵×0.355
V = 2.14065×10⁵ V.
V ≈ 2.14×10⁵ V.
amount of Work required to move a proton through the section is given as,
W = qV ............... Equation 2
Where W = work required to move a proton through the section, q = charge on a proton V = Electric potential.
Given: V = 2.14×10⁵ V, q = 1.60 x 10⁻¹⁹ C.
Substitute into equation 2
W = (2.14×10⁵)(1.60 x 10⁻¹⁹)
W = 3.424×10⁻¹⁴ J.
