A: capacitance C of system B: the potential difference between the charge center

Question

A: capacitance C of system B: the potential difference between the charge center and the ground C: average strength of the electric field between cloud and ground D: the electric energy stored in the system Problem 5 (10 POINTS) The charge center of a thundercloud, drifting 2.8km above the earth's surface, contains 24.0C of negative charge. Assuming the charge center has a radius of modeling the charge center and the earth's surface as parallel plates, calculate (a) the capacitance C of the system, (b) the potential difference between the charge center and the ground, (c) the average strength of the electric field between cloud and ground, (d) the electrical energy stored in the system. charge center has a radius of R = 1.0km, and

Explanation / Answer

distance between two plates d=2.8 km =2800 m

charge q=24 C

radius of the plare r=1km =1000m

a)

capacitance C=eo*A/d =eo*(pi*r^2)/d

C=8.85*10^-12*(pi*1000^2)/2800

C=9.93*10^-9 C

b)

potentail V=q/C

V=24/(9.93*10^-9)

V=2.42*10^9 V

c)

electric field E=V/d

E=(2.42*10^9)/(2800)

E=8.643*10^5 V/m

d)

enrgy stored U=1/2*C*v^2

U=1/2*(9.93*10^-9)*(2.42*10^9)^2

U=2.91*10^10 J

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