Question

Electric Potential (Parallel Plate Capacitor Potential Energy and Potential)

A parallel plate capacitor has two terminals, one (+) and the other (-). When you move a test positive charge, q at uniform velocity from the negative terminal (Ui and Vi) to the positive terminal (Uf and Vf), work W = ΔU = qΔV is done on the charge, increasing the energy of the field by this amount. The work done by the field on the charge is – W. (V = U/q, all have their usual meaning)

QUESTIONS on the above observations:

(i) What is the work done to move a negative charge, q at uniform velocity from the positive to the negative terminal? Increasing the potential energy, a push (work) is needed to move the object.

(ii) Which terminal is the high potential for the plus charge?

(iii) Which terminal is the high potential for the negative charge?

(iv) If you do work 1.0 eV to move a proton from the negative to the positive terminal of a capacitor, how much work will you do to move an electron in the exact same manner from the positive the negative terminal of the same capacitor?

(v) What is the potential difference across the above capacitor?

Answer 1

i)Work done to move a negative charge from positive to negative terminal W=qV [ That is work is same as in the case of positive charge as q and  V are same]

ii)Positive terminal is the high potential for positive charge . [As positive terminal repels the positive charge ,work is to be done to move towards it]

iii)Negative terminal is the high potential for negative charge . [As negative  terminal repels the negative charge, work is to be done to move towards it]

iv)As the work done to move proton  is 1eV, work done to move electron is also 1eV because proton and electron have same charge q=e and here potential difference is also same  

V . [work W=qV=eV]

v)Work W=eV (W=1eV)

1eV=eV

Hence potential difference V=1volt