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Standard XII
Physics
Question
The electric potential at a point
P
(
x
,
y
,
z
)
is given by
V
=
−
x
2
y
−
x
z
3
+
4
. The electric field
→
E
at that point is
(
2
x
y
+
z
3
)
^
i
+
x
2
^
j
+
3
x
z
2
^
k
2
x
y
^
i
+
x
2
^
j
+
3
x
z
2
^
k
(
2
x
y
+
z
3
)
^
i
+
(
x
2
+
y
2
)
^
j
+
(
3
x
y
−
y
2
)
^
k
2
x
y
−
z
3
^
i
+
x
y
2
^
j
+
3
z
2
x
^
k
A
2
x
y
^
i
+
x
2
^
j
+
3
x
z
2
^
k
B
(
2
x
y
+
z
3
)
^
i
+
x
2
^
j
+
3
x
z
2
^
k
C
(
2
x
y
+
z
3
)
^
i
+
(
x
2
+
y
2
)
^
j
+
(
3
x
y
−
y
2
)
^
k
D
2
x
y
−
z
3
^
i
+
x
y
2
^
j
+
3
z
2
x
^
k
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Solution
Verified by Toppr
V
=
−
x
2
y
−
x
z
3
+
4
→
E
=
−
∂
V
∂
x
^
i
−
∂
V
∂
y
^
j
−
∂
V
∂
z
^
k
Now,
∂
V
∂
x
=
−
2
x
y
−
z
3
∂
V
∂
y
=
−
x
2
∂
V
∂
z
=
−
3
x
z
2
Hence,
→
E
=
(
2
x
y
+
z
3
)
^
i
+
x
2
^
j
+
3
x
z
2
^
k
Answer-(A)
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Similar Questions
Q1
The electric potential at a point
P
(
x
,
y
,
z
)
is given by
V
=
−
x
2
y
−
x
z
3
+
4
. The electric field
→
E
at that point is
View Solution
Q2
The electric potential at a point
(
x
,
y
,
z
)
is given by
V
=
−
x
2
y
−
x
z
3
+
4
. The electric field
→
E
at that point is
View Solution
Q3
The electric field potential in space has the form
V
(
x
,
y
,
z
)
=
−
2
x
y
+
3
y
z
−
1
,. The electric field intensity
→
E
magnitude at the point (-1, 1 , 2) is
View Solution
Q4
Given
→
F
=
(
2
x
y
+
z
2
)
^
i
+
x
2
^
j
+
2
x
z
^
k
. Then
→
F
is :-
View Solution
Q5
The electric potential
V
at any point
P
(
x
,
y
,
z
)
in space is given by
V
=
4
x
2
V
. The electric field at the point
(
1
m
,
2
m
)
is:
View Solution