Let a point charge Q125 nC be located at P₁ (4, -2, 7) and a charge Q2 = 60 nC be at P2(-3, 4,-2). a) If = 0, find E at P3(1, 2, 3): This field will be 10-9 [25R13, 60R23 + LIR1313 R23³ [23 4760 where R13 = -3a, +4ay-4a- and R₂3 = 4a, -2ay +5a.. Also, R13 = √41 and [R₂3] =√45. So E 10-9 [25 x E= (-3ax +4ay -4a₂) 60 x (4a, -2ay +5a₂)] (41)1.5 + (45)1.5 [²5 × (- 4740 = 4.58a, -0.15ay +5.51a. b) At what point on the y axis is Ex=0? P3 is now at (0, y, 0), so R₁3 = -4a, + (y + 2)ay-7a. and R23 = 3ax +(y-4)ay +2a. Also, [R13] =√√65+ (y + 2)² and |R23] =√13+(y-4)². Now the x component of E at the new P3 will be: 25 × (-4) 10-⁹ 470 [65+(y+2)2]1.5 Ex + 0, we require the expression in the large brackets to be zero. This expression To obtain Ex = simplifies to the following quadratic: 60 x 3 [13+(y-4)²]¹.5 0.48y2 + 13.92y + 73.10 = 0 which yields the two values: y=-6.89, -22.11
Let a point charge Q125 nC be located at P₁ (4, -2, 7) and a charge Q2 = 60 nC be at P2(-3, 4,-2). a) If = 0, find E at P3(1, 2, 3): This field will be 10-9 [25R13, 60R23 + LIR1313 R23³ [23 4760 where R13 = -3a, +4ay-4a- and R₂3 = 4a, -2ay +5a.. Also, R13 = √41 and [R₂3] =√45. So E 10-9 [25 x E= (-3ax +4ay -4a₂) 60 x (4a, -2ay +5a₂)] (41)1.5 + (45)1.5 [²5 × (- 4740 = 4.58a, -0.15ay +5.51a. b) At what point on the y axis is Ex=0? P3 is now at (0, y, 0), so R₁3 = -4a, + (y + 2)ay-7a. and R23 = 3ax +(y-4)ay +2a. Also, [R13] =√√65+ (y + 2)² and |R23] =√13+(y-4)². Now the x component of E at the new P3 will be: 25 × (-4) 10-⁹ 470 [65+(y+2)2]1.5 Ex + 0, we require the expression in the large brackets to be zero. This expression To obtain Ex = simplifies to the following quadratic: 60 x 3 [13+(y-4)²]¹.5 0.48y2 + 13.92y + 73.10 = 0 which yields the two values: y=-6.89, -22.11
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(Electromagnetic field)I want a detailed solution for this solution
![5-
Let a point charge Q125 nC be located at P₁ (4, -2, 7) and a charge Q2 = 60 nC be at P2(-3, 4, -2).
a) If = 0, find E at P3(1, 2, 3): This field will be
10-9 [25R13
4760 LIR1313
√45.
So
where R13 = -3a, +4ay-4a. and R₂3 = 4a, -2ay +5a. Also, R13 = √√41 and [R23] =
10-9 [25 x (-3ax +4ay -4a₂) 60 x (4a, 2ay +5a₂)
(41)1.5
+
4740
(45)1.5
= 4.58a, -0.15ay + 5.51a.
6-
E=
7-
Ex =
E=
b) At what point on the y axis is Ex=0? P3 is now at (0, y, 0), so R₁3 = -4ax + (y + 2)ay-7a.
and R23 = 3ax +(y-4)ay +2a. Also, [R₁3] = √65+ (y + 2)² and [R23] =√13+ (y-4)².
Now the x component of E at the new P3 will be:
10-⁹
25
(-4)
x
400 [[65 45 (x+2)²31.5+ [13 +60 3271.5
To obtain Ex= 0, we require the expression in the large brackets to be zero. This expression
simplifies to the following quadratic:
which yields the two values: y=-6.89, -22.11
Ep =
+
0.48y2 + 13.92y + 73.10 = 0
Point charges of 120 nC are located at A(0, 0, 1) and B(0, 0, -1) in free space.
a) Find E at P(0.5, 0, 0): This will be
60R23
R23³
120 × 10-⁹ RAP
47€0
IRAP³
1.1x
0 - 10 11 1² fotot L² (²
where RAP = 0.5axa and Rgp = 0.5ax +az. Also, [RAP] = [RBP| = √1.25. Thus:
120 x 10-⁹ax
Ep =
4x (1.25)1.50
2-10 [-+20
=10(-3.39) (.0266) (.626)=0.57 C
=
+
sin
The region in which 4 <r < 5,0 <0 < 25°, and 0.97 << 1.17 contains the volume charge
density of p, 10(r-4)(r-5) sin 0 sin(p/2). Outside the region, p, = 0. Find the charge within
the region: The integral that gives the charge will be
RBP
RBP|³
772 V/m
(r-4) (r-5) sin 0 sin(/2) r² sin 0 dr do do
(20) [2005 (9)
-2 cos](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb910055f-6f5e-4012-8d6c-1cf81811edbd%2F95b8b37c-ea4d-4075-b324-60f6c8df077d%2Fss654xn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5-
Let a point charge Q125 nC be located at P₁ (4, -2, 7) and a charge Q2 = 60 nC be at P2(-3, 4, -2).
a) If = 0, find E at P3(1, 2, 3): This field will be
10-9 [25R13
4760 LIR1313
√45.
So
where R13 = -3a, +4ay-4a. and R₂3 = 4a, -2ay +5a. Also, R13 = √√41 and [R23] =
10-9 [25 x (-3ax +4ay -4a₂) 60 x (4a, 2ay +5a₂)
(41)1.5
+
4740
(45)1.5
= 4.58a, -0.15ay + 5.51a.
6-
E=
7-
Ex =
E=
b) At what point on the y axis is Ex=0? P3 is now at (0, y, 0), so R₁3 = -4ax + (y + 2)ay-7a.
and R23 = 3ax +(y-4)ay +2a. Also, [R₁3] = √65+ (y + 2)² and [R23] =√13+ (y-4)².
Now the x component of E at the new P3 will be:
10-⁹
25
(-4)
x
400 [[65 45 (x+2)²31.5+ [13 +60 3271.5
To obtain Ex= 0, we require the expression in the large brackets to be zero. This expression
simplifies to the following quadratic:
which yields the two values: y=-6.89, -22.11
Ep =
+
0.48y2 + 13.92y + 73.10 = 0
Point charges of 120 nC are located at A(0, 0, 1) and B(0, 0, -1) in free space.
a) Find E at P(0.5, 0, 0): This will be
60R23
R23³
120 × 10-⁹ RAP
47€0
IRAP³
1.1x
0 - 10 11 1² fotot L² (²
where RAP = 0.5axa and Rgp = 0.5ax +az. Also, [RAP] = [RBP| = √1.25. Thus:
120 x 10-⁹ax
Ep =
4x (1.25)1.50
2-10 [-+20
=10(-3.39) (.0266) (.626)=0.57 C
=
+
sin
The region in which 4 <r < 5,0 <0 < 25°, and 0.97 << 1.17 contains the volume charge
density of p, 10(r-4)(r-5) sin 0 sin(p/2). Outside the region, p, = 0. Find the charge within
the region: The integral that gives the charge will be
RBP
RBP|³
772 V/m
(r-4) (r-5) sin 0 sin(/2) r² sin 0 dr do do
(20) [2005 (9)
-2 cos
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