# Calculate the dipole moment of an electron and a proton that the electric field 75.0cm away has the magnitude 2.30N/C

Grade:12th pass

## 2 Answers

Samyak Jain
333 Points
4 years ago
Considering length of dipole to be very small i.e. much less than 75 cm
and the point at which electric field is given is on the axis of dipole, we have
r = 75 cm = 0.75 m ,  E = 2.30 N/C
E = 2 K P / r3 , where K = 1/4$\pi$$\varepsilon$0 = 9 x109 Nm2/C2 and P is dipole moment.
2.3 = (2 x 9 x109 x P) / (0.75)3  $\dpi{80} \Rightarrow$  P = (2.3)(0.75)3 / (2x9x109)
P = 0.0539 x 109 = 5.39 x 10–9 Cm
The required dipole moment is 5.39 x 10–9 Cm.
Samyak Jain
333 Points
4 years ago
A slight mistake is there in the above answer.
The correct value is P = 5.39 x 10 –11.
Ans. of attachment :
We know that force on a negative charge or acceleration of a negative charge is opposite to the electric field.
So, if the acceleration of electron is eastward, then the electric field is westward.
Let a, m and e be the acceleration, mass and charge of the electron,
F be the force due to electric field exerted on it and
E be the magnitude of the electric field. Then,
F = e.E  and  F = ma
e.E = ma  $\dpi{80} \Rightarrow$  E = ma / e
Putting the respective values, we get E = (9.1 x 10–31 kg)(1.84 x 109 m/s2) / (1.6 x 10–19 C)  (approx.)
E = 10.465 x 10–3 N/C = 1.0465 x 10–2 N/C
Thus, magnitude and direction of electric field is 1.0465 x 10–2 N/C and west respectively.

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