Q. 714.0( 2 Votes )

Two particles A and B, having opposite charges 2.0 × 10–6 C and –2.0 × 10–6 C, are placed at a separation of 1.0 cm.

(a) Write down the electric dipole moment of this pair.

(b) Calculate the electric field at a point on the axis of the dipole 1.0 m away from the center.

(c) Calculate the electric field at a point on the perpendicular bisector of the dipole and 1.0 m sway from the center.


Answer :


Given:
(a)
Charge on particle A : q1 = 2.0 × 10–6 C
Charge on particle B : q2 = -2.0 × 10–6 C
Magnitude of both the charges : q = 2.0 × 10–6 C
Separation between A and B : d = 1.0 cm = 0.01 m
(b)
Distance between center and the point on the axis of dipole: r= 1 cm= 0.01 m
(c)
Distance between center and the point on the perpendicular bisector of the dipole:
r’= 1 m
Formula used:
(a)
Electric dipole moment is given as:
Where, is the electric dipole moment, q is the magnitude of the charges at the end of the dipole and is the vector joining the two charges.


Hence, electric dipole moment between A and B is 2× 10-8Cm
(b)
Electric field at a point on the axis of the dipole is:
Here k is a constant and k= =9× 109 Nm2C-2, p is the Electric dipole moment and r is the distance between the point on the axis and it’s center.
Substituting we get,

Hence, electric field at a distance 1 cm away from the center of the dipole to the point on it’s axis is 3.6× 108 NC-1.
(c)
Electric field at a point on the perpendicular bisector of the dipole is given as:
Here, k is a constant and k= =9× 109 Nm2C-2, p is the Electric dipole moment and r’ is the distance between the point on the perpendicular bisector of the dipole and it’s center
Substituting we get:

Hence, electric field at a point on the perpendicular bisector of the dipole 1 m away from it’s center is 180 NC-1.


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