In the parallel-plate capacitor of Fig. 24.2, suppose the plates are pulled apart so that the separation d is much larger than the size of the plates.

(a) Is it still accurate to say that the electric field between the plates is uniform? Why or why not?

(b) In the situation shown in Fig. 24.2, the potential difference between the plates is V_ab = Qd/ε₀A. If the plates are pulled apart as described above, is V_ab more or less than this formula would indicate? Explain your reasoning.

(c) With the plates pulled apart as described above, is the capacitance more than, less than, or the same as that given by Eq. (24.2)? Explain your reasoning.

Fig.24.2

Eq.24.2

Transcribed Image Text:

(a) Arrangement of the capacitor plates Plate a, area A Wire Potential difference = Vab Plate b, area A Wire (b) Side view of the electric field E + ++ When the separation of the plates is small compared to their size, the fringing of the field is slight. Capacitance of a parallel-plate capacitor - in vacuum -Magnitude of charge on each plate A** Area of each plate . Distance between plates Vab Potential difference between plates Electric constant