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EAMCET Engineering 2003 Solved Paper
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© examsnet.com
Question : 19
Total: 160
An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along
X
-axis at
x
=
1
cm
,
x
=
2
cm
,
x
=
4
cm
,
x
=
8
cm
.
.
.
. and so on. In this set up if the consecutive charges have opposite sign, then the electric field in newton/coulomb at
x
=
0
is
:
(
1
4
π
ε
0
=
9
×
10
9
N
−
m
2
∕
C
2
)
12
×
10
4
24
×
10
4
36
×
10
4
48
×
10
4
Validate
Solution:
Electric field intensity due to a point charge.
E
=
1
4
π
ε
0
⋅
Q
r
2
The consecutive charges are of opposite signs.
∴
Net electric field at
x
=
0
, is
E
=
1
4
π
ε
0
[
Q
r
1
2
−
Q
r
2
2
+
Q
r
3
2
−
Q
r
4
2
.
.
.
.
.
.
∞
]
=
Q
4
π
ε
0
[
1
r
1
2
−
1
r
2
2
+
1
r
3
2
−
1
r
4
2
+
.
.
.
∞
]
=
9
×
10
9
×
5
×
10
−
9
[
1
(
1
×
10
−
2
)
2
−
1
(
2
×
10
−
2
)
2
+
1
(
4
×
10
−
2
)
2
+
.
.
.
∞
]
=
45
10
−
4
[
1
1
2
−
1
2
2
+
1
4
2
−
1
8
2
+
.
.
.
.
∞
]
=
45
×
10
4
[
1
1
2
−
1
2
2
+
1
4
2
−
1
8
2
.
.
.
∞
]
⌊
1
1
2
−
1
2
2
+
1
4
2
−
1
8
2
.
.
.
∞
]
=
1
1
2
1
−
(
−
1
2
)
2
[
Sum of infinite GP,
S
∞
=
a
1
−
r
]
=
1
1
+
1
4
=
4
5
∴
E
=
45
×
10
4
×
4
5
=
36
×
10
4
N
∕
C
© examsnet.com
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