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marissa [1.9K]
3 years ago
5

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)
Physics
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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b) Check explanatios for b

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b) y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}

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y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right]

A linear equation is of the form y = Gx

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x = \left[\begin{array}{ccc}100\end{array}\right]

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