Explain and answer please 1. Three point charges q1= 1mc, q2=2mc and q3=-3mc are

Question

Explain and answer please 1. Three point charges q1= 1mc, q2=2mc and q3=-3mc are respectively located at (0,0,4),(-2,6,1) and (3,-4,-8) 1.1 determine q1 1.2 find the electric field intensity at q1

2. Given that the current density j is given by j=10(x^2+y^2)az [A/m^2] 2.1 determine the current density at point (-3,4,6) 2.2 find the current I, crossing an area on the xy plane given by 0<=x<=1 , 0<=y<=3 Hint I=double integral (s) jds; ds=azdxdy

3.consider the conducting plates shown in fig if v(z=0)= 0, and v(z=5mm)=100v, determine 3.1 the electric potential V 3.2 the electric field intensity , E

Explanation / Answer

FOLLOW THIS a) let charge q1 = -4 nC , charge q2 = 5 nC and charge q3 = 3 nC the distance between point P and charge q1 is d1= 0.5 m +1.2 m = 1.7 m the distance between point P and charge q2 is d2 = 1.2 m the distance between point P and charge q3 is d3 = 1.2 - 0.8 = 0.4 m the electric field on point P due to charge q1 is E1 = k(q1)/(d1)2 E1 = [(9*109 N.m2/C2)(4*10-9 C)]/(1.7 m)2 E1 = 12.45 N/C this is along -ve x-direction, so, E1 = -(12.45 N/C) i^ the electric field on point P due to charge q2 is E2 = [(9*109 N.m2/C2)(5*10-9 C)]/(1.2 m)2 E2 = (31.25 N/C) i^ the electric field on point P due to charge q3 is E3 = [(9*109 N.m2/C2)(3*10-9 C)]/(0.4 m)2 E3 = (168.75 N/C) i^ toatal electric field E = E1 + E2 + E3 =-12.45 + 31.25 + 168.75 = 187.55 N/C i^ electric field at (1.2 , 0) is Ex = (187.55 N/C) i^ ................................................................................................ from fig, AB = 0.5 m , BC = 0.8 m and BP = 1.2 m from triangle ? PAB, AP = ?[(0.5)2 + (1.2)2] = 1.3 m from triangle ? PAB, sin?A = 1.2/1.3 = 0.9230 cos?A = 0.5/1.3 = 0.3846 from triangle ?PCB, PC = ?[(0.8)2 + (1.2)2] = 1.44 m from triangle ? PCB, sin?C = 1.2/1.44 = 0.8333 cos?C = 0.5/1.44 = 0.9996 x-components of the electric field : Ex = -EA sin?A i^ + EC cos?C i^ = [(9*109 )(4*10-9)/(1.3)2](0.3846) + [(9*109)(3*10-9)/(1.44)2](0.9996) = (21.2 N/C) i^ y-components of the electric field : Ey = EA sin?A j^ + EC sin?C j^+ EB j^ = -[(9*109 )(4*10-9)/(1.3)2](0.9230) + [(9*109)(3*10-9)/(1.44)2](0.8333) + [(9*109)(5*10-9)/(1.2)2] = (-19.66 + 10.85 + 31.25) = (22.44 N/C) j^ resultant electric field is E = (21.2 N/C) i^ + (22.44 N/C) j^ answer : E = (21.2 N/C) i^ + (22.44 N/C) j^ ................................................................................ note : magnitude of the electric field is E = ?[(Ex)2 + (Ey)2] = 30.87 N/C electric field at (0 , 1.2) is E = 30.87 N/

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