A parallel-plate capacitor has plates with area 0.0225 m2 separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 V. (b) Use Gauss’s law (Eq. 24.23) to calculate the electric field inside the Teflon. (c) Use Gauss’s law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.
Solution 46E Step 1 of 4: Gauss’s law in dielectrics has the same form as in vacuum except that the electric field is multiplied by a factor of K and the charge enclosed by the Gaussian surface is the free charge. The capacitance of an object depends on its shape and size . When the space between the conductors is filled with a dielectric material, the capacitance increases by a factor K called the dielectric constant of the material. The quantity = K 0 C = KC 0 Step 2 of 4: a)The capacitance of a parallel plate capacitor is 12 2 2 C = K0A = 2.1×8.85×10 C /N.m ×(0.0225)= 4.18 × 10 10F d 1×10 m The charge on its plates is Q = CV = 4.18 × 10 10F × 12 V