A proton is released from rest in a uniform electric field that has a magnitude of 8.0 x 104 V/m. The proton undergoes a displacement of 0.50 m in the direction of E Find the change in electric potential, the voltage, between point A and B. Find the change in potential energy of the proton-field system for this displacement. Hint: U

Explanation:

The magnitude of electric field = 8 x 10⁴ V /m

there is a potential difference of 8 x 10⁴ V on a separation of 1 m

so on a separation of .5 m , potential drop or change in potential will be equal to

.5 x 8 x 10⁴

= 4 x 10⁴ V .

The increase in kinetic energy of proton = V X Q

V is potential drop x Q is charge on proton

= 4 x 10⁴ x 1.6 x 10⁻¹⁹

= 6.4 x 10⁻¹⁵ J

potential energy of the proton-field system will be correspondingly decreased by the same amount or by an amount of

-  6.4 x 10⁻¹⁵ J .

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