NCERT Class XII Chapter
Electric Charges and Fields
Questions With Solutions
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Question : 28
Total: 34
(a) A conductor A with a cavity is shown in figure (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q. (figure (b))
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q. (figure (b))
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Solution:
(a) Let us consider a closed surface inside the conductor enclosing the cavity. As electric field inside the conductor is zero, so
ϕ =∮ s
.
By Gauss’s theorem,
ϕ =
By equations (i) and (ii), we found that
q enclosed
This shows that there cannot be any charge on the inner surface of conductor at the cavity, so any charge Q given to the conductor will appear only on its outer surface.
(b) Let us consider that the charge –q 1 q 1
Let us construct a closed Gaussian surface S inside the conductor A enclosing its cavity. As, electric field inside conductor is zero, so
ϕ =∮ s
.
But, by Gauss’s theorem
ϕ =
By equations (iii) and (iv), we get
q − q 1 q 1
i.e., equal and opposite charge, –q is induced on inner surface of conductor A at the cavity, when conductor B of charge, + q is kept is its cavity.
Hence, the charge on outer surface of conductor is Q +q 1
(c) As we found that electric field inside the cavity of conductor is zero, even on charging the conductor, so the sensitive instrument can be shielded from the strong electrostatic fields in its environment, by covering it with a metallic cover.
ϕ =
By Gauss’s theorem,
ϕ =
By equations (i) and (ii), we found that
This shows that there cannot be any charge on the inner surface of conductor at the cavity, so any charge Q given to the conductor will appear only on its outer surface.
(b) Let us consider that the charge –
Let us construct a closed Gaussian surface S inside the conductor A enclosing its cavity. As, electric field inside conductor is zero, so
ϕ =
But, by Gauss’s theorem
ϕ =
By equations (iii) and (iv), we get
i.e., equal and opposite charge, –q is induced on inner surface of conductor A at the cavity, when conductor B of charge, + q is kept is its cavity.
Hence, the charge on outer surface of conductor is Q +
(c) As we found that electric field inside the cavity of conductor is zero, even on charging the conductor, so the sensitive instrument can be shielded from the strong electrostatic fields in its environment, by covering it with a metallic cover.
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