8- Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density, 0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: 9- [(12 × 10-⁹) (-2ax - 6a-)] (3 x 10-9) (2a,- 3ay)] (4+36)1.5 + ]-[ 4740 2760 (4+9) = -3.9a, - 12.4a, - 2.5a, V/m E= Let E = 400ax - 300ay +500a, in the neighborhood of point P (6, 2, -3). Find the incremental work done in moving a 4-C charge a distance of 1 mm in the direction specified by: a) ax + ay+az: We write dW=-qE.dL = -4(400a, - 300ay +500a₂).. (4 x 10-³) √√3 dW=-qE.dL=-4(400ax - 300ay +500a-).. (4 x 10-³) √14 (400-300+500) = -1.39 J 1/E.. 3 b) -2ax +3ay-a: The computation is similar to that of part a, but we change the direction: (-2ax +3ay-a-) √14 (10-³) W = -9 =-6 -(-800-900-500) = 2.35 J 10- Find the amount of energy required to move a 6-C charge from the origin to P(3, 1, -1) in the field E = 2xax - 3y²ay + 4a, V/m along the straight-line path x = -3z, y = x + 2z: We set up the computation as follows, and find the the result does not depend on the path. -6/120 E.dL-6 (0.2 x 10-9)ax 2€0 (a, +ay+a-) √√3 1 · 6 " 3y²dy - 6 Jo 2xdx + 6 (10-³) of (2xax − 3y²ay +4a-) · (dxax +dyay+dza;) - 4dz = -24J
8- Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density, 0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: 9- [(12 × 10-⁹) (-2ax - 6a-)] (3 x 10-9) (2a,- 3ay)] (4+36)1.5 + ]-[ 4740 2760 (4+9) = -3.9a, - 12.4a, - 2.5a, V/m E= Let E = 400ax - 300ay +500a, in the neighborhood of point P (6, 2, -3). Find the incremental work done in moving a 4-C charge a distance of 1 mm in the direction specified by: a) ax + ay+az: We write dW=-qE.dL = -4(400a, - 300ay +500a₂).. (4 x 10-³) √√3 dW=-qE.dL=-4(400ax - 300ay +500a-).. (4 x 10-³) √14 (400-300+500) = -1.39 J 1/E.. 3 b) -2ax +3ay-a: The computation is similar to that of part a, but we change the direction: (-2ax +3ay-a-) √14 (10-³) W = -9 =-6 -(-800-900-500) = 2.35 J 10- Find the amount of energy required to move a 6-C charge from the origin to P(3, 1, -1) in the field E = 2xax - 3y²ay + 4a, V/m along the straight-line path x = -3z, y = x + 2z: We set up the computation as follows, and find the the result does not depend on the path. -6/120 E.dL-6 (0.2 x 10-9)ax 2€0 (a, +ay+a-) √√3 1 · 6 " 3y²dy - 6 Jo 2xdx + 6 (10-³) of (2xax − 3y²ay +4a-) · (dxax +dyay+dza;) - 4dz = -24J
8- Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density, 0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: 9- [(12 × 10-⁹) (-2ax - 6a-)] (3 x 10-9) (2a,- 3ay)] (4+36)1.5 + ]-[ 4740 2760 (4+9) = -3.9a, - 12.4a, - 2.5a, V/m E= Let E = 400ax - 300ay +500a, in the neighborhood of point P (6, 2, -3). Find the incremental work done in moving a 4-C charge a distance of 1 mm in the direction specified by: a) ax + ay+az: We write dW=-qE.dL = -4(400a, - 300ay +500a₂).. (4 x 10-³) √√3 dW=-qE.dL=-4(400ax - 300ay +500a-).. (4 x 10-³) √14 (400-300+500) = -1.39 J 1/E.. 3 b) -2ax +3ay-a: The computation is similar to that of part a, but we change the direction: (-2ax +3ay-a-) √14 (10-³) W = -9 =-6 -(-800-900-500) = 2.35 J 10- Find the amount of energy required to move a 6-C charge from the origin to P(3, 1, -1) in the field E = 2xax - 3y²ay + 4a, V/m along the straight-line path x = -3z, y = x + 2z: We set up the computation as follows, and find the the result does not depend on the path. -6/120 E.dL-6 (0.2 x 10-9)ax 2€0 (a, +ay+a-) √√3 1 · 6 " 3y²dy - 6 Jo 2xdx + 6 (10-³) of (2xax − 3y²ay +4a-) · (dxax +dyay+dza;) - 4dz = -24J
(Electromagnetic field)Hello sir I want a detailed solution for this solution
Transcribed Image Text:8-
Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC
at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density,
0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is:
9-
[(12 x 10-⁹) (-2ax – 6az)
4740
(4+36)1.5
= -3.9a, 12.4a, - 2.5a, V/m
E=
==
Let E = 400ax - 300ay +500a; in the neighborhood of point P (6, 2, -3). Find the incremental work
done in moving a 4-C charge a distance of 1 mm in the direction specified by:
a) ax + ay+az: We write
+
dW=-qE-dL=-4(400a, - 300ay +500a₂)..
(4 x 10-³)
√3
==
dW=-qE dL = -4(400ax - 300ay +500a-)..
(4 x 10-³)
√14
W = -9
(3 x 10-) (2a,- 3ay)
27€0
(4+9)
SE
(400-300+500) = -1.39 J
b) -2ax +3ay-a: The computation is similar to that of part a, but we change the direction:
(-2ax +3ay-a-)
√14
(10-³)
-3
-6/²221
2]-[
(-800-900-500) = 2.35 J
(0.2 x 10-9)ax
240
E.dL-6
10-
Find the amount of energy required to move a 6-C charge from the origin to P(3, 1, −1) in the field
E = 2xax - - 3y²ay + 4a V/m along the straight-line path x = -3z, y = x + 2z: We set up the
computation as follows, and find the the result does not depend on the path.
(ax + ay + a)
√√3
=-6 2xdx + 6
(10-³)
−6 [ (2xax − 3y¹²ay + 4ª-) · (dxax +dyay+dza:)
of ³y³²dy - 65
-6
4dz = -24J
Interaction between an electric field and a magnetic field.
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![8-
Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC
at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density,
0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is:
9-
[(12 x 10-⁹) (-2ax – 6az)
4740
(4+36)1.5
= -3.9a, 12.4a, - 2.5a, V/m
E=
==
Let E = 400ax - 300ay +500a; in the neighborhood of point P (6, 2, -3). Find the incremental work
done in moving a 4-C charge a distance of 1 mm in the direction specified by:
a) ax + ay+az: We write
+
dW=-qE-dL=-4(400a, - 300ay +500a₂)..
(4 x 10-³)
√3
==
dW=-qE dL = -4(400ax - 300ay +500a-)..
(4 x 10-³)
√14
W = -9
(3 x 10-) (2a,- 3ay)
27€0
(4+9)
SE
(400-300+500) = -1.39 J
b) -2ax +3ay-a: The computation is similar to that of part a, but we change the direction:
(-2ax +3ay-a-)
√14
(10-³)
-3
-6/²221
2]-[
(-800-900-500) = 2.35 J
(0.2 x 10-9)ax
240
E.dL-6
10-
Find the amount of energy required to move a 6-C charge from the origin to P(3, 1, −1) in the field
E = 2xax - - 3y²ay + 4a V/m along the straight-line path x = -3z, y = x + 2z: We set up the
computation as follows, and find the the result does not depend on the path.
(ax + ay + a)
√√3
=-6 2xdx + 6
(10-³)
−6 [ (2xax − 3y¹²ay + 4ª-) · (dxax +dyay+dza:)
of ³y³²dy - 65
-6
4dz = -24J](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2030af63-5d4a-4849-a910-42f05d92e55e%2Fe4395931-128e-4ede-8657-277a04438366%2F2zq88hq_processed.jpeg&w=3840&q=75)
