Consider a charged particle q0 with a charge of +0.001C and a mass of 2X10^-8 kg

Question

Consider a charged particle q0 with a charge of +0.001C and a mass of 2X10^-8 kg moving between conductors. b) what is the work done on q0 as it moves from point A to point B? Be sure to consider the magnitude and sign. C) Now if q0 is released from rest at point B what is the speed when it reaches point A? D) what is the work done on q0 as it moves from point C to point D? be sure to consider magnitude and sign E) what is the magnitude of the work done on q0 if it were to move between the points of maximum potential difference?

Explanation / Answer

(b)

Potential at point B, Vb = 0, since it is the neutral point

Potential at A, Va = -1 V

So, work done in moving charge from A to B is :

W = Potential difference*Magnitude of charge = (Vb-Va)*q0 = 1*0.001 = 0.001 J

(c)

If B is released and it moves towards A, then the work calculated in previous part will become its kinetic energy on reaching A

So, 1/2*m*v2 = 0.001

Putting m = 2*10-8 kg, we get : v = 316.22 m/s

(d)

Since both C and D lie on the same equipotential surface, so Vc-Vd = 0

Thus work done is zero

(e)

According to the given figure, maximum possible potential difference is : 2.5-(-2.5) = 5 V

So, maximum possible work done = 5*q0 = 5*0.001 = 0.005 J

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