Phy question ...electricity B

(a)(i) Charge density = (3.5 x 10^-15)/(4/3)pi.(2.31)^3 C/cm^3
Charge enclosed by sphere of radius 1.45 cm
= (3.5x10^-15) x (1.45/2.31)^3 C = 8.656 x 10^-16 C
Increase in potential at radius 1.45 cm
= (9x10^9).(8.656x10^-16)/0.0145 v = 5.373 x 10^-4 v
Hence potential = (5.373 x 10^-4 + 4x10^-4) v = 9.373 x 10^-4 v

(ii) Repeat the claculation using radius of 2.31 cm instead of 1.45 cm

(b)(i) Potential at A = (9x10^9).[-5/0.15 + 2/0.05]x10^-6 v = 60 000 v

(ii) Potential at B = (9x10^9).[-5/0.05 + 2/0.15] x 10^-6 v = -780 000 v

(iii) Work done = (3x10^-6).(60000 + 780000) J = 2.52 J

(iv) Increase

(v) Same work is done.



2013-05-04 23:18:47 補充：
sorry...the above answer is wrongly posted. The correct answers are briefly given below:
(a) Use E = (1/2)CV^2
dE/E = 2(dV/V)
0.1 = 2(dV/V)
dV/V = 0.05, i.e. 5%
V needs to be increased by 5%

2013-05-04 23:35:18 補充：
(b)(i) Charge density p = 8.4x10^-7/100x10-4 C/m^2 = 8.4x10^-5 C/m^2
1.4x10^6 = 8.4x10^-5/k(8.85x10^-12) where k is the dielectric constant
k = 6.78
(ii) E-field without dielectric = 8.4x10^-5/8.85x10^-12 v/m = 9.49x10^6 v/m
Induced charge = 8.4x10^-7 x (1 - 1/6.78) C = 7.16x10^-7 C

2013-05-04 23:41:49 補充：
(c) Use V = kQ/R, where Q is the charge on sphere surface, R is phere radius
i.e. Q = VR/k
E field at surface = kQ/R^2 = V/R = 8000/0.05 v/m = 160000 v/m
Energy density = (1/2).(8.85x10^-12).(160000)^2 J = 0.113 J