Question

30% Three very large planes carrying uniform surface charge densities are located in a medium with &r = 2 as shown in Figure 1. Draw the net electric field (E-field) due to the system. Explain what principles you used to obtain the net E-field and comment on the graph. Comment on any assumptions and approximations used. (ii) 15% Calculate the electric field strength and displacement field at the three points shown in Figure 1. 64 =-10 nC/m202 = 10 nC/m² 02=-10 nC/m2 P, PP 20 cm 20 cm 10 cm Figure 1 (iii) 40% Two point charges, qı = 5 nC and q2 = -4 nC, are located in free space at (0, 0.5, 0.5) m and (0.1, 0.7, 0.7) m respectively in the vicinity of a conducting grounded surface. The surface lies in the (x, y) plane at z = 0 as shown in Figure 2. Find the electric potential at (0, 0.1, 0.1) m. Comment on the method you used to obtain the answer. The reference electric potential is (0) = 0. (iv) 15% Find the electric flux due to the system described in part (111) of this question passing through a spherical surface of radius 1 m centred on the origin.

Answer 1

Three very large planes carrying unipom surface charge densitice are located as shown in below fig. = Lonc./m? 52= lonc/m² 193= -loncima. 20cm Tloom. We know that, Electric field intensity by .. Large planes is given by I re it is. independent of distance from the surface... Therefore, het electric field intensities at different locations are , . Ex=2. fo Enett = -10 - 460. | Erett - 10 Enetl=10) Enett = 10 460. 46o. * * tonam² lonc/m² -lone/m2. r : Electric field goes away from the charge - Surface density and coming towards to CSscanned with CamScanteie change surface density 7

* Xlet electric field due to the system is Assumption large planes. Therefore, no convex error at the corner of plane • Magnitude of electic field intensity every fehere is same but direction is different. UN Ep, a le az ; Ephen ar '; EP, bene Similory, De = ar DP = 10 x ; Des = 10 az CS$canned with CamScanner

(iii) conducting ground surface at agy) plane. Two point charges q = she and q2 = -4nc are located at (0, 0.5, 0.5) and (0.1,0.7₂0.7 respectively. - Here, we will use concept of mirror image of charge. Since, surface present is conducting and to satisfy boundary conditions, two image charges are present at (0,0.5, -0.5) and 0.1, 0.1, -0.7) efe charge - snc & 4nc respectively. Therefore, electric potential at (0,0.1,0.1) is given by Knott = 4 + V₂ + V₂ + vy. K snc. - Snc -4nc I J (0-4)2 + 0.412 10.1)2+0.6)2 +6.6)2 (0.6)2 the head over 101:] Wohere K = utta = 9X 109 Nm2. 4TGO cz . csscanned with CamScanner

V = 915 _ 5 0:72 0.57 0.85 = 9 [8.77 – 6:94 - 4.7 +47. 7 Kit. Vrett = 10.17 V According to Gauss' law e. closed surface Closed surface - Venclosed to. Since, due to presence of conducting surface, virtual image charge will come. Therefore, het charge enclosed the side sphere of radius im is o. Therefore, Flux = 07** CSScanned with CamScanner