A proton is released from rest in a uniform electric field that has a magnitude of $8.0 \times 10^{4} V / m$ (Fig. 25.6). The proton undergoes a displacement of 0.50 m in the direction of E.

(A) Find the change in electric potential between points A and B.

(B) Find the change in potential energy of the proton–field system for this displacement.

(C) Find the speed of the proton after completing the 0.50 m displacement in the electric field.

Question

A proton is released from rest in a uniform electric field that has a magnitude of [latex] 8.0 imes 10^{4} V / m [/latex] (Fig. 25.6). The proton undergoes a displacement of 0.50 m in the direction of E.

(A) Find the change in electric potential between points A and B.

(B) Find the change in potential energy of the proton–field system for this displacement.

(C) Find the speed of the proton after completing the 0.50 m displacement in the electric field.

Accepted Answer

Because the positively charged proton moves in the direction of the field, we expect it to move to a position of lower electric potential. From Equation 25.6, we have

[latex] \Delta V=-E \int_{A}^{B} d s=-E d [/latex] (25.6).

[latex] \Delta V=-E d=-\left(8.0 imes 10^{4} V / m ight)(0.50 m )=-4.0 imes 10^{4} V [/latex].

(B) Find the change in potential energy of the proton–field system for this displacement.

Using Equation 25.3,

[latex] \Delta V \equiv \frac{\Delta U}{q_{0}}=-\int_{A}^{B} E \cdot d s [/latex] (25.3)

[latex] \Delta U=q_{0} \Delta V=e \Delta V [/latex].

[latex] =\left(1.6 imes 10^{-19} C ight)\left(-4.0 imes 10^{4} V ight) [/latex].

[latex] =-6.4 imes 10^{-15} J [/latex].

The negative sign means the potential energy of the system decreases as the proton moves in the direction of the electric field. As the proton accelerates in the direction of the field, it gains kinetic energy and at the same time the system loses electric potential energy.

(C) Find the speed of the proton after completing the 0.50 m displacement in the electric field.

The charge–field system is isolated, so the mechanical energy of the system is conserved:

[latex] \Delta K+\Delta U=0 [/latex].

[latex] \left(\frac{1}{2} m v^{2}-0 ight)+e \Delta V=0 [/latex].

[latex] v=\sqrt{\frac{-(2 e \Delta V)}{m}} [/latex].

[latex] =\sqrt{\frac{-2\left(1.6 imes 10^{-19} C ight)\left(-4.0 imes 10^{4} V ight)}{1.67 imes 10^{-27} kg }} [/latex].

[latex] =2.8 imes 10^{6} m / s [/latex].

Question 25.2: A proton is released from rest in a uniform electric field t...

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