In Fig. 26-32, a battery of potential difference V = 27 V is connected to a resi

Question

In Fig. 26-32, a battery of potential difference V = 27 V is connected to a resistive strip of resistance R = 9.1 . When an electron moves through the strip from one end to the other, (a) in which direction in the figure does the electron move (1 = upward, 2 = downward), (b) how much work (in eV) is done on the electron by the electric field in the strip, and (c) how much energy (in eV) is transferred to the thermal energy of the strip by the electron?

(a)

Number

Units

(b)

Number

Units

(c)

Number

Units

Explanation / Answer

(a) in which direction in the figure does the electron move (1 = upward, 2 = downward), 

I  can't see the drawing however electron would move from the negative terminal of the battery, throught the strip towards the negative terminal. This will be apposite to the direction of conventional current ( from positive sourse terminal, to the load and finally returning to the  negative terminal)

(b)how much work (in eV) is done on the electron by the electric field in the strip,  

Work doneW  by a potential difference V  on charge q is 

W=qV  since we have an electron q= 1.602 x10-19 C

W=1.602 x10-19 x  27= 43.2x10-19 J or since 1J = 6.24x1018 eV 

W= 43.2x10-19x 6.24x1018 = 27 eV

(c) how much energy (in eV) is transferred to the thermal energy of the strip by the electron?

We know that work done on electron by moving across the strip is W = 27 eV 

This is equivalent if was moved  in vacum. We also know that by moving a bunch of electrons(charge)  per unit of time is power P = VI or P= V2/R; Power is also heat/time. P=E/t then thermal energy E is 

E= (V2/R)/t = V2/(Rt)  if we know the everage drif velocity v of an electron and length L of the strip we can compute time t as t=L/v

the drift velocity can be computed from the current I equation 

I = n q A v  where

n - number of free electrons (depends on a conductor)

q= 1.602 x10-19 C

A - area of crossection of the strip 

v - drift velocity 

I -current and also I=V/R

v= I/ n q A = V/(n q A R)

Now E=  V2/(Rt) becomes

E=V2v/(RL) = V2 V/(n q A R2L) ) =V3  /(n q A R2L) )

Please provide edditional information to solve this problem.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects