how I can do this A cross-section of equipotential surfaces is shown below: The

Question

how I can do this A cross-section of equipotential surfaces is shown below: The surfaces/lines containing the points a, b, c & d have electric potentials of 600 volts, 200 volts, & -300 volts respectively: Determine / calculate the following: The magnitude of the Electric Field along the regions be where distance be is 2 cm. The work (in eV) required to move an electron from point b to d: The work (in eV) required to move a proton from point b to a: The work (in eV) required to move am electron from point d to c:

Explanation / Answer

The magnitude of electric field is E =V/d

a)

The magnitude of the electric field along the region bc is given at distance d =2cm =0.02m is

E =200-(-300)/0.02 =25000=25kV/m

b)

The work done to move an electron from point b to d is given by

W =qV =-1*200-(-300) =-500*1.6*10-19/1.6*10-19 =-500eV   ( we know that 1eV =1.6*10-19 J)

c)

The work done to remove a proton from b to a is given by W =-1*(200-600) =-1*-400 =400eV

d)

The work done to remove an electron from the point d to c is given by

W =qV =-1*0 =0 (zero) because the potential difference is zero

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