A proton with an initial speed of 200,000 m/s moves 1.5 m in the same direction as a uniform electric field of E = 750 N/C points. What is the change in electric potential energy of the proton in Joules along its path?
Added by Juan Antonio H.
Step 1
Given: Electric field strength (E) = 750 N/C, charge of proton (q) = 1.6 x 10^-19 C The force (F) experienced by the proton in the electric field is given by the equation F = q * E. Substitute the values: F = 1.6 x 10^-19 C * 750 N/C = 1.2 x 10^-16 N. Show more…
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