4- Let Q1 = 8 μC be located at P₁ (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let € = 0. a) Find F2, the force on Q2: This force will be Q122 R12 F2 = 470 R1213 (8 x 10-6)(-5 x 10-6) (4ax +10ay) 47 €0 (116)1.5 (-1.15ax -2.88ay) mN b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be: Q3 Q1R13 Q₂R23 + 470 R1313 R233 where R₁3 = (x - 2)a, + (y – 5)ay and R₂3 = (x − 6)ax + (y - 15)ay. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q₂. The slope of this vector is (15 — 5)/(6-2) = 2.5. Therefore, we look for P3 at coordinates (x, 2.5.x, 8). With this restriction, the force becomes: F3 = or Q38[(x-2)ar +2.5(x - 2)ay] 5[(x-6)a, +2.5(x - 6)ay] F3 = 47 €0 [(x-2)²+(2.5)²(x − 2)²]¹.5¯¯ [(x − 6)² + (2.5)²(x − 6)²]¹.5 where we require the term in large brackets to be zero. This leads to which reduces to - 8(x − 2)[((2.5)² + 1)(x − 6)²]¹.5 — 5(x − 6)[((2.5)² + 1)(x − 2)²]¹5 = 0 - = 8(x − 6)²-5(x - 2)² = 0 - 6√8-2√5 √8-√5 The coordinates of P3 are thus P3 (21.1, 52.8, 8) X = = 21.1
4- Let Q1 = 8 μC be located at P₁ (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let € = 0. a) Find F2, the force on Q2: This force will be Q122 R12 F2 = 470 R1213 (8 x 10-6)(-5 x 10-6) (4ax +10ay) 47 €0 (116)1.5 (-1.15ax -2.88ay) mN b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be: Q3 Q1R13 Q₂R23 + 470 R1313 R233 where R₁3 = (x - 2)a, + (y – 5)ay and R₂3 = (x − 6)ax + (y - 15)ay. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q₂. The slope of this vector is (15 — 5)/(6-2) = 2.5. Therefore, we look for P3 at coordinates (x, 2.5.x, 8). With this restriction, the force becomes: F3 = or Q38[(x-2)ar +2.5(x - 2)ay] 5[(x-6)a, +2.5(x - 6)ay] F3 = 47 €0 [(x-2)²+(2.5)²(x − 2)²]¹.5¯¯ [(x − 6)² + (2.5)²(x − 6)²]¹.5 where we require the term in large brackets to be zero. This leads to which reduces to - 8(x − 2)[((2.5)² + 1)(x − 6)²]¹.5 — 5(x − 6)[((2.5)² + 1)(x − 2)²]¹5 = 0 - = 8(x − 6)²-5(x - 2)² = 0 - 6√8-2√5 √8-√5 The coordinates of P3 are thus P3 (21.1, 52.8, 8) X = = 21.1
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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(Electromagnetic field) I want a detailed solution for this solution
![4-
Let Q1 = 8 μC be located at P₁ (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let e = €0.
a) Find F2, the force on Q2: This force will be
Q1 Q2 R12
F2 =
470 R12³
b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in
general will be:
Q1R13 Q2R23
+
R133 R23³
where R₁3 = (x - 2)a¸ + (y – 5)ay and R23 = (x − 6)ax + (y - 15)ay. Note, however, that
all three charges must lie in a straight line, and the location of Q3 will be along the vector R12
extended past Q₂. The slope of this vector is (15 — 5)/(6-2) = 2.5. Therefore, we look for P3
at coordinates (x, 2.5.x, 8). With this restriction, the force becomes:
F3
or
(8 × 10-6)(−5 × 10−6) (4ax + 10ay)
(116)1.5
47 €0
=
F3 =
which reduces to
Q3
470
where we require the term in large brackets to be zero. This leads to
Q3
8[(x-2)a, - 2)ay]
5[(x-6)a, +2.5(x - 6)ay]
47€0 [[(x - 2)² + (2.3)²(x - 2)²3³-³3- ((x - 5²4 (23) ²(x - 621-3
8(x − 2)[((2.5)² + 1)(x − 6)²]¹.5 – 5(x − 6)[((2.5)² + 1)(x − 2)²]¹-5 = 0
8(x − 6)² - 5(x - 2)² = 0
-
(-1.15a, -2.88ay) mN
X =
6√8-2√5
√8-√5
The coordinates of P3 are thus P3 (21.1, 52.8, 8)
= 21.1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb910055f-6f5e-4012-8d6c-1cf81811edbd%2F19ac03e7-1d0d-4675-b7f2-d02b0cecbe87%2F3kygbrh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4-
Let Q1 = 8 μC be located at P₁ (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let e = €0.
a) Find F2, the force on Q2: This force will be
Q1 Q2 R12
F2 =
470 R12³
b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in
general will be:
Q1R13 Q2R23
+
R133 R23³
where R₁3 = (x - 2)a¸ + (y – 5)ay and R23 = (x − 6)ax + (y - 15)ay. Note, however, that
all three charges must lie in a straight line, and the location of Q3 will be along the vector R12
extended past Q₂. The slope of this vector is (15 — 5)/(6-2) = 2.5. Therefore, we look for P3
at coordinates (x, 2.5.x, 8). With this restriction, the force becomes:
F3
or
(8 × 10-6)(−5 × 10−6) (4ax + 10ay)
(116)1.5
47 €0
=
F3 =
which reduces to
Q3
470
where we require the term in large brackets to be zero. This leads to
Q3
8[(x-2)a, - 2)ay]
5[(x-6)a, +2.5(x - 6)ay]
47€0 [[(x - 2)² + (2.3)²(x - 2)²3³-³3- ((x - 5²4 (23) ²(x - 621-3
8(x − 2)[((2.5)² + 1)(x − 6)²]¹.5 – 5(x − 6)[((2.5)² + 1)(x − 2)²]¹-5 = 0
8(x − 6)² - 5(x - 2)² = 0
-
(-1.15a, -2.88ay) mN
X =
6√8-2√5
√8-√5
The coordinates of P3 are thus P3 (21.1, 52.8, 8)
= 21.1
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