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3 years ago

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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.

What is the electric field strength at the points...
A) (x,y) = (12.0 cm, 0 cm)
B) (x,y) = (0 cm, 12.0 cm)

1 answer:

Ronch [10]3 years ago

7 0

Answer:

Value of electric field along the axis and equitorial axis $E=31.25\ N/c$ and $E = 15.625\ N/c$ respectively.

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) :

Electric field of dipole in its axis ,

$E=\dfrac{2kp}{r^3}$

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

We get , $E=31.25\ N/c.$

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

Electric field of dipole on equitorial axis ,

$E = \dfrac{kp}{r^3}$

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

We get , $E = 15.625\ N/c.$

Hence , this is the required solution.

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