Can someone please help with this problem? Thanks II. Parallel plate Capacitor C
ID: 1998388 • Letter: C
Question
Can someone please help with this problem? Thanks II. Parallel plate Capacitor Consider the triangular parallel plate capacitor below in which the total charge on the 6.00 uC. The separation between the plates is d 2.00 mm. Let the height plates is Q 15.0 mm and the baseb 25.0 mm. 10. Determine the electric field E between the plates. 11. Determine the potential difference AV between the plates. 12. Determine the surface charge density o on the plates 13. What is the capacitance of the plates? 14. If a dielectric of dielectric constant K is placed entirely between the plates, what is the new capacitance? 15. What is the physical significance of capacitance that is nor obvious from the definition of capacitance? front view side view
Explanation / Answer
Here,
Q = 6 uC
seperation , d = 2 mm = 0.002 m
capacitance , C = 8.854 *10^-12 * 0.5 * 15 *10^-3 * 25 *10^-3/0.002
C = 8.3 *10^-13 F
10)
electric field = V/d
electric field = (6 *10^-6/8.3 *10^-16)/(2 *10^-3)
electric field = 3.61 *10^12 N/C
11)
potential difference = 3.61 *10^24 * 0.002
potential difference = 7.23 *10^9 V
12)
suraface charge density = electric field * epsilon
suraface charge density = 3.61 *10^12 * 8.854 *10^-12
suraface charge density = 31.96 C/m^2
13)
Capacitance = 8.3 *10^-13 F
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please post others in a seperate question