How much work does the electric field do in moving a proton from a point with a potential of +V1 = +185 V to a point where it is V2 = 55.0 V?

W=  2.08 x 10⁻¹⁷  J

Explanation:

Given that

Initial potential V₁ =185 V

Final potential V₂ = 55 V

We know that charge of the proton

q=1.6 x 10⁻¹⁹ C

Work done is given as

W= q ΔV

q=Charge

ΔV=Potential difference

W=Work done

Now by putting the values in the above equation then we get

W= 1.6 x 10⁻¹⁹  ( 185 - 55 ) J

W=208 x 10⁻¹⁹ J

W=  2.08 x 10⁻¹⁷  J

Therefore the work done will be 2.08 x 10⁻¹⁷  J.

answer:

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