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TS EAMCET 04 May 2019 Shift 1 Solved Paper
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© examsnet.com
Question : 105
Total: 160
Anon-conducting solid sphere has radius R and uniform charge density. A spherical cavity of radius R/4 is hollowed out of the sphere. The distance between center of sphere and center of cavity is R/2. If the charge of the sphere is Q after the creation of the cavity and the magnitude of electric field at the center of the cavity is E = K
(
Q
4
π
∊
0
R
2
)
, determine the approximate value of K
0.32
0.78
0.51
0.45
Validate
Solution:
Consider the below diagram
The expression for the electric field is written as,
E
=
E
Q
0
−
E
cavity
=
Q
0
4
π
ε
0
R
2
(
R
2
)
−
0
=
Q
0
4
π
ε
0
R
2
(
1
2
)
.
.
.
(
1
)
The expression for the charge after the creation of the cavity is calculates as,
Q
Q
0
=
4
3
π
(
R
3
−
(
R
4
)
3
)
4
3
π
R
3
For uniform charge density the ratio of the charge is given by,
Q
Q
0
=
63
64
Q
0
=
63
64
Q
Substitute
63
64
Q
for
Q
0
in the equation (1)
E
=
Q
4
π
ε
0
R
2
(
1
2
)
=
Q
4
π
ε
0
R
2
(
64
63
×
1
2
)
=
0.51
Q
4
π
ε
0
R
2
This implies
k
=
0.51
© examsnet.com
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