Two large parallel conducting plates separated by 9 cm carry equal and opposite

Question

Two large parallel conducting plates separated by 9 cm carry equal and opposite surface chargedensities such that the electric field between them is uniform. Thedifference in potential between the plates is 200 V. An electron is released from rest at thenegative plate. (a) What is the magnitude of the electric fieldbetween the plates?
1 kV/m
Which plate is at the higher potential? 2 The negative plate
the positiveplate    


(b) Find the work done by the electric field on the electron as theelectron moves from the negative plate to the positive plate.Express your answer in both electron volts and joules.
3 J
4 eV

(c) What is the change in potential energy of the electron when itmoves from the negative plate to the positive plate?
5 eV
What is its kinetic energy when it reaches the positive plate?
6 eV (a) What is the magnitude of the electric fieldbetween the plates?
1 kV/m
Which plate is at the higher potential? 2 The negative plate
the positiveplate    


(b) Find the work done by the electric field on the electron as theelectron moves from the negative plate to the positive plate.Express your answer in both electron volts and joules.
3 J
4 eV

(c) What is the change in potential energy of the electron when itmoves from the negative plate to the positive plate?
5 eV
What is its kinetic energy when it reaches the positive plate?
6 eV 2 The negative plate
the positiveplate    
2 The negative plate
the positiveplate    
2

Explanation / Answer

Here distance (d) between the plates =9 cm =0.09 m Also potential difference (PD)=200 V a)We know , electric field = Potential difference / distance =200/0.09 = 2222.22 V/m = 2.22 kV/m    Here positive plate is at higher potential . b)Also work done = Potential difference * charge = 200 V *1e =200eV =200*1.6*10-19 J = 3.2*10-17 J c)Change in potential energy = Potential difference * charge =200V*-1e = -200 eV d)From law of conservation of energy ,    decrease in potential energy = increase in kineticenergy    So increase in kinetic energy =200 eV

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