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3 years ago

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A charge of 19 nC is uniformly distributed along a straight rod of length 15 m that is bent into a circular arc with a radius of

5.7 m. What is the magnitude of the electric field at the center of curvature of the arc?

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

The magnitude of the electric field at the center of curvature of the arc is 3.87 N/C

Explanation:

Please see the attachments below

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An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.

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Explanation:

Given :

Distance between charges , $d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.$

Magnitude of charges , $q=1\ nC = 10^{-9}\ C.$

Dipole moment , $p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.$

Case A) (x,y) = (12.0 cm, 0 cm) :

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Putting all values and $r=12\times 10^{-2}\ m.$

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Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

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Putting all values and $r=12\times 10^{-2}\ m.$

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