A uniform electric field of magnitude 400 N/C pointing in the positive x-directi

Question

A uniform electric field of magnitude 400 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 2.80 cm.

The work done by the field on the electron is -1.792*10^-18 J

The change in potential energy associated with the electron is 1.792*10^-18 J

The velocity of the electron is 1.97*10^6 m/s

A proton is now fired in the same direction from the same position.what is the change in potential energy associated with the proton if he moved the same distance (2.80 cm)?

Explanation / Answer

change in Potential energy, deltaPE = q deltaV

and deltaPE = - Work done by field

deltaV = - E.d

= - (400)(2.80/100)

= - 11.2 Volt

change in PE = (1.6 x 10^-19)(-11.2)

= - 1.792 x 10^-18 J ......Ans

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