A point charge $$+ q$$ is placed at mid-point of a cube of side $$L.$$ The electric flux emerging from the cube is
A.
$$\frac{q}{{{\varepsilon _0}}}$$
B.
$$\frac{{6q{L^2}}}{{{\varepsilon _0}}}$$
C.
$$\frac{q}{{6{L^2}{\varepsilon _0}}}$$
D.
zero
Answer :
$$\frac{q}{{{\varepsilon _0}}}$$
Solution :
By Gauss’s theorem, total electric flux over the closed surface is $$\frac{1}{{{\varepsilon _0}}}$$ times the total charges contained inside surface.
∴ Total electric flux $$ = \frac{{{\text{total charge inside cube}}}}{{{\varepsilon _0}}}$$
or $$\phi = \frac{q}{{{\varepsilon _0}}}$$
Releted MCQ Question on Electrostatics and Magnetism >> Electric Field
Releted Question 1
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
Two point charges $$ + q$$ and $$ - q$$ are held fixed at $$\left( { - d,o} \right)$$ and $$\left( {d,o} \right)$$ respectively of a $$x-y$$ coordinate system. Then
A.
The electric field $$E$$ at all points on the $$x$$-axis has the same direction
B.
Electric field at all points on $$y$$-axis is along $$x$$-axis
C.
Work has to be done in bringing a test charge from $$\infty $$ to the origin
D.
The dipole moment is $$2qd$$ along the $$x$$-axis
Three positive charges of equal value $$q$$ are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in
A uniform electric field pointing in positive $$x$$-direction exists in a region. Let $$A$$ be the origin, $$B$$ be the point on the $$x$$-axis at $$x = + 1cm$$ and $$C$$ be the point on the $$y$$-axis at $$y = + 1cm.$$ Then the potentials at the points $$A,B$$ and $$C$$ satisfy: