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along the axis of a dipole, 0.5 m from its cen- ter, you measure an electric field of magni- tude 30 n/c. the charge separation between the two point charges in the dipole is 0.004 m. what is the charge at the positive end of the dipole?

Sagot :

bharathparasad577

The resulting dipole is 0.56 N/C

Charges of the dipole, q = 30

Separation distance between the charges, d = 0.0010 m

Separation distance between the center and the charge, d' = d/2 = 0.25

x = 4.0 m

y = 0.0 m

Now,

The electric field due to the positive charge on the right of the origin:

E = k{q/(d'+x)²}

where

k = Coulomb's constant = 9 x 10⁹ Nm²/C²

Now,

E = 1124.72 N/C

Similarly, electric field due to the negative charge:

E' = k{q/(x-d')²}

E' = -1125.72 N/C

Thus

E(total) = E' - E = 0.56 N/C

Therefore, the positive charge is 0.56 N/C.

To know more about charge, refer: https://brainly.com/question/14592329

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