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Light from the sun allows a solar cell to move electrons from the positive to the negative terminal, doing 2.4×10-19Jof work per electron. What is the emf of this solar cell?

Short Answer

Expert verified

The emf of the solar cell is1.5V

Step by step solution

01

Given information and theory used

Given :

Work per electron : 2.4×10-19J

Theory used :

The work is equal toW=q∆V

02

Calculating the emf of the solar cell 

Writing ∆Vasemf,εin our case, we have :

ε=Wq=2.4×10-191.6×10-19=1.5V

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Most popular questions from this chapter

The parallel-plate capacitor in Figure Q26.11 is connected to a battery having potential difference ∆Vbat. Without breaking any of the connections, insulating handles are used to increase the plate separation to 2d.

a. Does the potential difference ∆VCchange as the separation increases? If so, by what factor? If not, why not?

b. Does the capacitance change? If so, by what factor? If not, why not?

c. Does the capacitor charge Qchange? If so, by what factor? If not, why not?

A typical cell has a membrane potential of -70mV, meaning that the potential inside the cell is 70mV less than the potential outside due to a layer of negative charge on the inner surface of the cell wall and a layer of positive charge on the outer surface. This effectively makes the cell wall a charged capacitor. Because a cell's diameter is much larger than the wall thickness, it is reasonable to ignore the curvature of the cell and think of it as a parallel-plate capacitor. How much energy is stored in the electric field of a 50μmdiameter cell with a 7.0nmthick cell wall whose dielectric constant is 9.0?

a. Use the methods of Chapter 25 to find the potential at distance xon the axis of the charged rod shown in FIGURE P26.43.

b. Use the result of part a to find the electric field at distance xon the axis of a rod

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