Question

6. 0.25/1 points Previous Answers SerPSET9 26.P.009.MI. My Notes An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 17.0-V potential difference is applied to these plates. (a) Calculate the electric held between the plates. 10'14 Remember that the electric field is uniform between the plates. What is the relationship between the electric field and electric potential in this case? kV/m (b) Calculate the surface charge density 8.854'10"-8 X There are a number of ways to get to this answer. How do you calculate the electric field due to a uniform surface charge density ncm (c) Calculate the capacitance. 3.96 pf (d) Calculate the charge on each plate. 6.73'10" 11 X You know the potential difference and the capacitance from part (c). How are these two quantities related to the charge on the capacitor pc Need Help?
can someone get the right answers. i dont wanna waste my tries so pls get it right. circle each answer pls and thank u. CLEAR TO SEE

Answer 1

t Air filled Capacitors- Area =7.60 cm² Seberation = 1.7 mm potential. = 170 4 + + min. Stay Cafecitore, C= KAGE Where Kal for air medium Lo, C= 7.6 x 10.07) * x 8-854 x1012- 1-7 X10-3 = 3.96 X1012 F = 3.96 RF 2 : 0 Slectric field = E= = y y los = 107 motfm. Lunface change dermation :P , kain, Este o= 10txt kako = 101XKGO = 107 X 1X 8854 x 10-12 . - ģ: 854 X10? unit = 88.5 very Mozart = 3.96 X17X10"= geesam 76X104 unit

(1) Charge on each plate, Q1 = OxA = 8859X108 x 7.6X104. = 6.73.X1011 Coulombs. = 67.3 pe A By Qaev = 3,96 x 17 . = 67.3 pe... A