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2 years ago

8

The potential difference between two points is 100 V. If a particle with a charge of 2 C is transported from one of these points

to the other, what is the magnitude of the work done?

2 answers:

Sunny_sXe [5.5K]2 years ago

7 0

1 volt = 1 joule/coulomb

100 volts = 100 joules/coulomb

Work = (100 J/C) x (2 C)

Work = (+ or -) 200 J

The work could be positive or negative. It depends on direction of the potential difference, direction the charge is moved, sign of the charge, and whether we mean work done by the mover or by the potential difference.

But the MAGNITUDE of the work is 200J .

Aleonysh [2.5K]2 years ago

4 0

Answer: 200 J

Explanation: In order to explain this we have consider that the work done in a electric field is given by:

Work= Q*ΔV=2*100=200J

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The magnitude of F₁ is 3.7 times of F₂

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Given that,

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For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

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$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

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Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

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