20)
Define electric field at a point. Is it a scalar or a vector quantity? Why should
the test charge qo be infinitesimally small for finding the electric field at a
point? Show that the electric field due to a small electric dipole at a point on the
perpendicular bisector or broad-side on position, is inversely proportional to
the third power of the perpendicular distance between the point and the line
joining the charge.
Answers
Answer:
Explanation:
The Field of an Electric Dipole
Electric field produced by a dipole is known as dipole field.
Let +q and -q be equal and opposite point charges separated by a small distance 2l. The strength of an electric dipole is measured by a vector quantity known as electric dipole moment known as electric dipole moment
, which is the product of the charge and separation between the charges, that is
The direction of
is always from negative to positive. The SI unit of dipole movement is Coulomb-meter.
For points on axial line
The axial line of a dipole is the line passing through the positive and negative charges of the electric dipole.
Consider a system of charges (-q and +q) separated by a distance 2a. Let 'P' be any point on an axis where the field intensity is to be determined.
Electric field at P (EB) due to +q
Electric field at P due to -q (EA)
Net field at P is given by
Simplifying, we get
As a special case :
(b) For points on the equatorial line
An equatorial line of a dipole is the line perpendicular to the axial line and passing through a point mid way between the charges.
Electric Field Intensity due to a Dipole at a Point lying on the Perpendicular Bisector of a Dipole.
Consider a dipole consisting of -q and +q separated by a distance 2a. Let P be a point Consider a point P on the equatorial line.
The resultant intensity is the vector sum of the intensities along PA and PB. EA and EB
can be resolved into vertical and horizontal components. The vertical compents of EA and EB cancel each other as they are equal and oppositely directed. It is the horizontal components which add up to give the resultant field.
E = 2E
A
cos q
As 2qa = p
As a special case,
We find that at very far off points i.e., 2a < r.< r.
Electricity intensity at an axial point is twice the electric intensity on the equatorial line.
The formula is correct if used with vector notation as the direction of electric dipole moment is from -q(negative charge) to +q(positive charge) and the resultant electric field at any point on equatorial line is in direction opposite to the electric dipole moment so negative sign is used. The other formula which does not include negative sign is applicable to calculate the magnitude(not direction) of electric field.