11.
A point charge $$+Q$$ is positioned at the center of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is
Flux going in pyramid $$ = \frac{Q}{{2{\varepsilon _0}}}.$$
which is divided equally among all 4 faces.
∴ Flux through one face $$ = \frac{Q}{{8{\varepsilon _0}}}$$
12.
An electric dipole is put in north-south direction in a sphere filled with water. Which statement is correct?
A
Electric flux is coming towards sphere.
B
Electric flux is coming out of sphere.
C
Electric flux entering into sphere and leaving the sphere are same.
D
Water does not permit electric flux to enter into sphere.
Answer :
Electric flux entering into sphere and leaving the sphere are same.
If electric dipole, the flux coming out from positive charge is equal to the flux coming in at negative charge i.e. total charge on sphere = 0. From Gauss law, total flux passing through the sphere = 0.
13.
A hollow metal sphere of radius $$5 cms$$ is charged such that the potential on its surface is $$10\,volts.$$ The potential at the centre of the sphere is
A
zero
B
$$10\,volts$$
C
same as at a point $$5 cms$$ away from the surface
D
same as at a point $$25 cms$$ away from the surface
The potential at the surface of a hollow or conducting sphere is same as the potential at the centre of the sphere and any point inside the sphere.
14.
A uniformly charged and infinitely long line having a linear charge density $$\lambda $$ is placed at a normal distance $$y$$ from a point $$O.$$ Consider an imaginary sphere of radius $$R$$ with $$O$$ as centre and $$R > y.$$ Electric flux through the surface of the sphere is
A
zero
B
$$\frac{{2\lambda R}}{{{\varepsilon _0}}}$$
C
$$\frac{{2\lambda \sqrt {{R^2} - {y^2}} }}{{{\varepsilon _0}}}$$
D
$$\frac{{\lambda \sqrt {{R^2} + {y^2}} }}{{{\varepsilon _0}}}$$
As we know expression of Gauss’s law is $$\oint {E \cdot ds = \frac{{{q_{{\text{inside}}}}}}{{{\varepsilon _0}}}} $$
$${\text{So,}}\,\,\phi = \frac{q}{{{\varepsilon _0}}}\,\,\left[ {\phi = \oint {E \cdot ds} } \right]$$
This is the net flux coming out from cube.
Since, a cube has 6 sides so electric flux through any face is $$\phi ' = \frac{\phi }{6} = \frac{q}{{6{\varepsilon _0}}}$$
17.
A charge is situated at a certain distance from an electric dipole in the end-on position experiences a force $$F.$$ If the distance of the charge is doubled, the force acting on the charge will be
$$E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{2p}}{{{r^3}}};E \propto \frac{1}{{{r^3}}} \Rightarrow F \propto \frac{1}{{{r^3}}}$$
Hence, the force will become $$\frac{F}{8}.$$
18.
The insulation property of air breaks down when the electric field is $$3 \times {10^6}V{m^{ - 1}}.$$ The maximum charge that can be given to a sphere of diameter $$5\,m$$ is approximately
Potential gradient relates with electric field according to the following relation, $$E = \frac{{ - dV}}{{dr}}$$
$$\eqalign{
& E = \frac{{ - dV}}{{dr}} \cr
& {\text{As,}}\,\,V = - {x^2}y - x{z^3} + 4 \cr
& {\text{So,}}\,\,E = - \frac{{dV}}{{dx}}\hat i - \frac{{dV}}{{dy}}\hat j - \frac{{dV}}{{dz}}\hat k \cr
& E = \left( {2xy + {z^3}} \right)\hat i + {x^2}\hat j + 3x{z^2}\hat k \cr} $$
20.
A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is
Let us consider a uniformly charged solid sphere without any cavity. Let the charge per unit volume be $$\sigma $$ and $$O$$ be the centre of the sphere. Let us consider a uniformly charged sphere of negative charged density $$\sigma $$ having its centre at $$O'.$$ Also let $$OO'$$ be equal to $$a.$$
Let us consider an arbitrary point $$P$$ in the small sphere. The electric field due to charge on big sphere
$${\overrightarrow E _1} = \frac{\sigma }{{3{\varepsilon _0}}}\overrightarrow {OP} $$
Also the electric field due to small sphere
$$\eqalign{
& {\overrightarrow E _2} = \frac{\sigma }{{3{\varepsilon _0}}}\overrightarrow {PO'} \,\,\,\,\therefore \,{\text{The total electric field}} \cr
& \overrightarrow E = {\overrightarrow E _1} + {\overrightarrow E _2} = \frac{\sigma }{{3{\varepsilon _0}}}\left[ {\overrightarrow {OP} + \overrightarrow {PO} } \right] = \frac{\sigma }{{3{\varepsilon _0}}}\overrightarrow {OO'} \cr} $$
Thus electric field will have a finite value which will be uniform.