The work done by an external force to move a particle of charge -2.20 uC at...

Question

Question

Answer 1

Answer #1

a)

From work energy

work done = change in potential energy

therefore,

the difference between the electric potential energy of charge at two points =6.40*10^-3 J

b)

delta K = Kf - Ki

but Vf=Vi

so, delta K = 0 J

c)

delta U =q*delta V

delta V= delta U/(q)

= 6.40*10^-3 J/(2.20*10^-6 C)

= 2909.090909 volts

delta V = 2909 volts

d)

point A is at higher potential than B.Since it requires positive work to move the negative charge from A to B.

Answer 2

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