parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference between the plates be to produce an energy density of 0.037 J/m3?

V = 576 V

Explanation:

Given:

- The area of the two plates A = 0.070 m^2

- The space between the two plates d = 6.3 mm

- Te energy density u = 0.037 J / m^3

Find:

- What must the potential difference between the plates V?

Solution:

- The energy density of the capacitor with capacitance C and potential difference V is given as:

u = 0.5*ε*E^2

- Where the Electric field strength E between capacitor plates is given by:

E = V / d

Hence,

u = 0.5*ε * (V/d) ^2

Where, ε = 8.854 * 10^-12

V^2 = 2*u*d^2 / ε

V = d*sqrt (2*u / ε)

Plug in values:

V = 0.0063*sqrt (2 * 0.037 / (8.854 * 10^-12))

V = 576 V