Work Done on A Proton in an Electric Field

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The equation for work done on a proton in an electric field is given as W = -qEd, resulting in a calculation of -2.76 x 10^-18 J. The confusion arises regarding why the work is reported as positive, despite the proton moving in the direction of the electric field. The key point is that the work done by the electric field is considered positive, as it indicates energy supplied to the proton. The negative sign in the equation reflects the work done against the field when considering the proton's charge. Ultimately, the positive value represents the energy gained by the proton from the electric field.
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Homework Statement
Calculate the work done in moving a proton 0.75 m in the same direction as the electric field
with a strength of 23 N/C
Relevant Equations
W=-qEd
W=-qEd
=-(1.6*10^-19)(23)(0.75)
= -2.76*10^-18 J

However, the answer is 2.76*10^-18 J. Why is the word done positive and not negative? Since it's traveling in the same direction as the electric field, shouldn't it be negative work?
 
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It seems to me the result quotes the work done by the field
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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