Consider two points in an electric field. The potential at point 1, V_1, is 27 V

Question

Consider two points in an electric field. The potential at point 1, V_1, is 27 V. The potential at point 2, V_2, is 166 V. A proton is moved from point 1 to point 2. Write an equation for the change of electric potential energy Delta U of the proton, in terms of the symbols given and the cha of the proton e. Solve for the numerical value of the change of the electric potential energy in electron volts (eV). Delta u = 2.224 * 10^-17 How much work is done by the electric force during the motion of the proton in J?

Explanation / Answer

Part (a):

Expression for change in electric potential energy ,

delta U = (V2 - V1)*e

Part (b):

Numerical value of electric potential energy in eV = (166 - 27) eV

= 139 eV

Part (c)

Work done during this movement = change in electric potential energy

= 139 eV = 139 x 1.6 x 10^-19 = 2.22 x 10^-17 J

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