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Test Index
JEE Main 29 Jan 2025 Shift 1 Paper
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Section:
Physics
Share question:
© examsnet.com
Question : 39 of 75
Marks:
+1
,
-0
List-I
List-II
(A)
Electric field inside (distance
r
>
0
r>0
r
>
0
from center) of a uniformly charged spherical shell with
surface charge density
σ
\sigma
σ
, and radius
R
R
R
.
(I)
σ
ε
0
\frac{\sigma}{\varepsilon_0}
ε
0
σ
(B)
Electric field at distance
r
>
0
r>0
r
>
0
from a uniformly charged infinite plane sheet with surface charge density
σ
\sigma
σ
.
(II)
σ
2
ε
0
\frac{\sigma}{2\varepsilon_0}
2
ε
0
σ
(C)
Electric field outside (distance
r
>
0
r>0
r
>
0
from center) of a uniformly charged spherical shell with
surface charge density
σ
\sigma
σ
, and
radius
R
R
R
.
(III)
0
(D)
Electric field between oppositely charged infinite plane parallel sheets with uniform
surface charge density
σ
\sigma
σ
.
(IV)
σ
ε
0
r
2
\;\frac{\sigma}{\varepsilon_0 r^2}
ε
0
r
2
σ
Choose the correct answer from the options given below:
[29 Jan 2025 Shift 1]
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
Validate
Solution:
👈: Video Solution
Inside uniformly charged spherical shell,
E
=
0
E=0
E
=
0
∴
A
→
I
I
I
\therefore \;\; A \rightarrow III
∴
A
→
III
For uniformly charged infinite plate
E
=
σ
2
ε
0
\;E=\;\frac{\sigma}{2\varepsilon_0}
E
=
2
ε
0
σ
B
→
I
I
\;B \rightarrow I I
B
→
II
Outside of spherical shell
E
=
Q
4
π
ε
0
r
2
=
σ
R
2
ε
0
r
2
E=\;\frac{Q}{4\pi\varepsilon_0 r_2}=\;\frac{\sigma R^2}{\varepsilon_0 r^2}
E
=
4
π
ε
0
r
2
Q
=
ε
0
r
2
σ
R
2
None of the option is matching for
C
C
C
.
Between two plates
E
=
σ
ε
0
E=\;\frac{\sigma}{\varepsilon_0}
E
=
ε
0
σ
D → I
None of the option is correct
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