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Byju's Answer
Standard XII
Physics
Potential at a Point Due to a Dipole
Find out the ...
Question
Find out the magnitude of electric field intensity and electric potential due to a dipole of moment
→
P
=
^
i
+
√
3
^
j
kept at origin at following points.
(i)
(
2
,
0
,
0
)
(ii)
(
−
1
,
√
3
,
0
)
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Solution
we have,
E
=
E
1
+
E
2
for axial dipole,
E
1
=
2
k
p
r
3
for equatorial diploe,
E
2
=
k
p
r
3
r
(
2
,
0
,
0
)
=
2
i
E
1
=
2
k
×
1
2
3
=
k
4
i
E
2
=
k
√
3
2
3
=
√
3
8
j
E
=
√
k
2
16
+
3
k
2
64
=
√
7
8
k
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0
Similar questions
Q.
Find out the magnitude of electric field intensity at point
(
2
,
0
,
0
)
due to a dipole of dipole moment
→
P
=
^
i
+
√
3
^
j
kept at origin. Also find the potential at that point.
Q.
Find out the magnitude of electric field intensity,
E
at point
(
2
,
0
,
0
)
due to a dipole moment,
→
p
=
^
i
+
√
3
^
j
kept at origin as shown in the figure and also find the potential,
V
at that point.
[
k
=
1
4
π
ε
0
]
Q.
Find out the magnitude of electric field intensity and electric potential due to a dipole of dipole moment
¯
P
=
^
i
+
√
3
^
j
kept at origin at following points.
(i) (2, 0, 0)
(ii) (-1,
√
3
, 0)
Q.
The magnitude of electric field intensity at point
P
(
2
,
0
,
0
)
due to a dipole of dipole moment,
→
p
=
^
i
+
√
3
^
j
kept at origin is -
(
Assume that the point
P
is at large distance from the dipole and
K
=
1
4
π
ϵ
0
)
Q.
An electric dipole of moment
→
p
=
(
^
i
−
3
^
j
+
2
^
k
)
×
10
−
29
Cm
, is at the origin. The electric field due to this dipole at
→
r
=
+
^
i
+
3
^
j
+
5
^
k
(note that
→
r
⋅
→
p
=
0
)
is parallel to:
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