Answer to Question #85778 in Electricity and Magnetism for Robert

Question

The parallel plates of an air capacitor are seperated by 2.25 mm. Each plate carries a charge of 6.5 nC. The magnitude of the electric field of the plates is 4.75x10⁵ V/m. Find the a) potential difference between the plates, b) capacitance, and c) area of a plate.

Answer 1

a) By definition of the potential difference is work on moving point charge between two plates. So

$$\Delta\phi=E\cdot d= 4.75\cdot10⁵ V/m \cdot 0.00225 m =1068.75 V$$

b)capacitance

$$C= \frac{Q}{\Delta\phi}=\frac{6.5 nC}{1068.75V}=6.08187\cdot10^{-12} F$$

c )

$$S=\frac{C\cdot d}{\varepsilon}=\frac{6.08187\cdot10^{-12} F\cdot 0.00225 m}{8.854F/m}=0.0015 m^2$$

Answer

$$\Delta\phi=1068.75 V$$ $$C=6.08187\cdot10^{-12} F$$

$$S=0.0015 m^2$$