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calculate the magnitude of an electric field due to an electric dipole of dipole moment 3.56*10^-29 at apoint 25.4nm away along the bisector axis |
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Answer» Answer: Dipole MOMENT =4×10 −11 ㎝ 9=4nC each distance between charges in dipole =d 9d=4×10 −9 ×d
=4×10 −11
d=10 −9 m a) At axis E due
to+9= (0.02) 2
K 9
E due
to−9= (0.02) 2
−K 9
E net
=K 9
[ (0.01) 2
1
− (0.02) 2
1
] =27×10 4 n/C At PERPENDICULAR 0.02m E net
=2n ∣E +9
∣=∣E −9
∣ at point P given y=0.02 E netP
=2∣E∣cosθ = ⎝ ⎛
( 2 d
) 2 +y 2
⎠ ⎞
2㎏
⋅ ⎝ ⎛
y 2 +( 2 d
) 2
⎠ ⎞
y
= ⎝ ⎛
( 2 0.01
) 2 +(0.02) 2
⎠ ⎞
2×9×10 9 ×4×10 −9
⋅ ⎝ ⎛
(0.02) 2 +( 2 0.01
) 2
⎠ ⎞
0.02
= (4.25×10 −4 ) 3/2
8
=9.13×10 5 N/C Explanation: hope it will HELP you...☺️ |
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