write the expression for the potential at any point due to an electric dipole and explain the terms
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get write the expression for the potential at any point due to an electric dipole and explain the terms from screen.
Write the expression for electric potential at a point due to an electric dipole and hence obtain the expression for the same at any point
Write the expression for electric potential at a point due to an electric dipole and hence obtain the ... axis and any point on its equatorial plane.
Write the expression for electric potential at a point due to an electric dipole and hence obtain the expression for the same at any point
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Derive an expression for electric potential at a point due to an electric dipole. Discuss the special cases.
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Derive an expression for electric potential at a point due to an electric dipole. Discuss the special cases.
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Updated on : 2022-09-05
Solution Verified by Toppr
Let an electric dipole consist of two equal and opposite point charges –q at A and +q at B ,separated by a small distance AB =2a ,with centre at O.
The dipole moment, p=q×2a
We will calculate potential at any point P, where
OP=r and ∠BOP=θ Let BP=r 1 and AP=r 2
Draw AC perpendicular PQ and BD perpendicular PO
In ΔAOC cosθ=OC/OA=OC/a
OC=acosθ
Similarly, OD=acosθ
Potential at P due to +q=
4πϵ 0 1 r 2 q
And Potential at P due to −q=
4πϵ 0 1 r 1 q
Net potential at P due to the dipole
V= 4πϵ 0 1 ( r 2 q − r 1 q ) ⟹V= 4πϵ 0 q ( r 2 1 − r 1 1 ) Now, r 1 =AP=CP =OP+OC =r+acosθ And r 2 =BP=DP =OP–OD =r−acosθ V= 4πϵ 0 q ( r−acosθ 1 − r+acosθ 1 ) = 4πϵ 0 q ( r 2 −a 2 cos 2 θ 2acosθ ) = r 2 −a 2 cos 2 θ pcosθ (Since p=2aq)
Special cases:-(i) When the point P lies on the axial line of the dipole, θ=0
If a<∘ cosθ=1 V= r 2 −a 2 p
r 2 p
Thus due to an electric dipole ,potential, V∝
r 2 1
(ii) When the point P lies on the equatorial line of the dipole, θ=90
∘ cosθ=0
i.e electric potential due to an electric dipole is zero at every point on the equatorial line of the dipole.
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Derive an expression for electric potential at point due to an electric dipole.Discuss the special cases.
Derive an expression for electric potential at point due to an electric dipole.Discuss the special cases. . Ans: Hint: An electric dipole is a pair of two charges equal in magnitude (q) but opposite in nature (i.e. one is a positive charge and other ...
Derive an expression for electric potential at point due to an electric dipole.Discuss the special cases.
Last updated date: 15th Mar 2023
• Total views: 236.7k • Views today: 4.32k Answer Verified 236.7k+ views 17 likes
Hint: An electric dipole is a pair of two charges equal in magnitude (q) but opposite in nature (i.e. one is a positive charge and other is a negative charge). The two charges are separated by a distance of length 2a.
Formula used: V= Kq r V=Kqr (1±x) n ≈1±nx (1±x)n≈1±nx
, if x is very much small.
Complete step by step answer:
An electric dipole is a pair of two charges equal in magnitude (q) but opposite in nature (i.e. one is a positive charge and other is a negative charge). The two charges are separated by a distance of length 2a.
An electric dipole has a quantity called its dipole moment given by P=2qa.
A sketch of an electric dipole is shown below.
Let an electric dipole with charges +q and –q lie on x-axis with the origin as the midpoint of the dipole.
Consider a point P at the coordinates (x,y). The distance of the point from origin be r making an angle
θ θ
with positive x-axis.
Electric potential due to a charge q at a point, which is at a distance r from the charge is given as
V= Kq r V=Kqr
, Where V is the potential due to the charge and K is permittivity of free space. Now, the electric potential at point P will be due two charges (+q and -q).
Let the potential due to charge +q be
V 1 V1 .
Let the potential due to charge -q be
V 2 V2 .
Let the total electric potential due to both the charges be V.
Since, electric potential is a scalar quantity,
V= V 1 + V 2 V=V1+V2 .
If you see the given figure,
V 1 = Kq r 1 V1=Kqr1 ………………(i).
Since, is right-angled triangle,
r 1 2 = (x−a) 2 + y 2 r12=(x−a)2+y2 . ⇒ r 1 = (x−a) 2 + y 2 − − − − − − − − − − − √ ⇒r1=(x−a)2+y2
Substitute the value of
r 1 r1 in equation (i). Therefore, V 1 = Kq (x−a) 2 + y 2 − − − − − − − − − − − √ V1=Kq(x−a)2+y2 . And V 2 =− Kq r 2 V2=−Kqr2 ………. (ii).
Is also a right-angled triangle. Therefore,
r 2 2 = (x+a) 2 + y 2 r22=(x+a)2+y2 . ⇒ r 2 = (x+a) 2 + y 2 − − − − − − − − − − − √ ⇒r2=(x+a)2+y2
Substitute the value of
r 2 r2 in equation (ii). Therefore, V 2 = −Kq (x+a) 2 + y 2 − − − − − − − − − − − √ V2=−Kq(x+a)2+y2 . This implies that V= V 1 + V 2 = Kq (x−a) 2 + y 2 − − − − − − − − − − − √ − Kq (x+a) 2 + y 2 − − − − − − − − − − − √
V=V1+V2=Kq(x−a)2+y2−Kq(x+a)2+y2
V=Kq ⎛ ⎝ ⎜ 1 (x−a) 2 + y 2 − − − − − − − − − − − √ − 1 (x+a) 2 + y 2 − − − − − − − − − − − √ ⎞ ⎠ ⎟
V=Kq(1(x−a)2+y2−1(x+a)2+y2)
…….(iii).
Consider the expression
(x±a) 2 + y 2 − − − − − − − − − − − √ (x±a)2+y2 .
Open up the brackets.
⇒ x 2 ±2ax+ a 2 + y 2 − − − − − − − − − − − − − − − √ ⇒x2±2ax+a2+y2 . …………... (1) But x 2 + y 2 = r 2 x2+y2=r2 .
Therefore, expression (1) can be written as
r 2 ±2ax+ a 2 − − − − − − − − − − − √ r2±2ax+a2 . Since r>>>>a, r 2 ±2ax+ a 2 − − − − − − − − − − − √ ≈ r 2 ±2ax − − − − − − − √ r2±2ax+a2≈r2±2ax
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