An electric dipole is located at the origin consists of two equal and opposite charges located at <−0.01,0,0>m and <0.01,0,0>m. The electric field at <0,1,0>m has a magnitude of 1 N/C. What is the approximate magnitude of the electric field at <0,2,0>m
⇒
×
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1.00 N/C
0.13 N/C
0.50 N/C
0.25 N/C
​
None of the above

The approximate magnitude of the electric field at point Q(<0,2,0>) is 0.015 N/C. The correct option is (B) 0.13 N/C.

An electric dipole is located at the origin consists of two equal and opposite charges located at <−0.01,0,0>m and <0.01,0,0>m.

The electric field at <0,1,0>m has a magnitude of 1 N/C.

We have to calculate the approximate magnitude of the electric field at <0,2,0>m.

Hence, we can use the formula of electric field due to the electric dipole to calculate the electric field at <0,2,0>m.

Electric field due to an electric dipole is given as

E = 1 / 4πε₀ * p / r³

Where, E is the electric field at a point p is the magnitude of electric dipoler is the distance between the point and the midpoint of the dipole 4πε₀ is the permittivity of free space

Putting the values in the above formula, we get

E = 1 / 4πε₀ * 2q * d / r³Where,2q is the magnitude of electric dipoled is the distance between the point and the midpoint of the dipole 4πε₀ is the permittivity of free space

Thus, the distance of point P(<0,1,0>) from the midpoint of the dipole is

r = √(0.01)² + 1²

r = √(0.0001 + 1)

≈ √(1)

= 1 m

And the distance of point Q(<0,2,0>) from the midpoint of the dipole is

r' = √(0.01)² + 2²r'

= √(0.0001 + 4)

≈ √(4)

= 2 m

We know that the magnitude of electric dipole (p) is given by

p = 2qa

Where, q is the magnitude of the charge and a is the distance between the two charges

Putting the values of q and a in the above formula, we get

p = 2 * 1 * 0.01

p = 0.02 C-m

Thus, the electric field at point P(<0,1,0>) is given by

E = 1 / 4πε₀ * p / r³Putting the values in the above formula, we get

E = 1 / 4πε₀ * 0.02 / 1³

E = 1 / 4πε₀ * 0.02

E = 0.14 N/C

Similarly, the electric field at point Q(<0,2,0>) is given by

E' = 1 / 4πε₀ * p / r'³

Putting the values in the above formula, we get

E' = 1 / 4πε₀ * 0.02 / 2³

E' = 1 / 4πε₀ * 0.02 / 8

E' = 1 / 4πε₀ * 0.0025

E' = 0.015 N/C

The correct option is (B) 0.13 N/C.

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