In building a particle accelerator, you manage to produce a uniform electric field of magnitude 6.03 × 10 5 N/C in one 35.5 cm section. Calculate the magnitude of the electric potential difference across the length of the accelerator's section. How much work is required to move a proton through the section?

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.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm