A square surface of side $$L$$ metre in the plane of the paper is placed in a uniform electric field $$E\left( {\frac{V}{m}} \right)$$ acting along the same place at an angle $$\theta $$ with the horizontal side of the square as shown in figure. The electric flux linked to the surface in unit of $$V$$ - $$m,$$ is
A.
$$E{L^2}$$
B.
$$E{L^2}\cos \theta $$
C.
$$E{L^2}\sin \theta $$
D.
$$0$$
Answer :
$$0$$
Solution :
Flux of electric field $$E$$ through any area $$A$$ is defined as $$\phi = E \cdot A\cos \theta $$ or $$\phi = E \cdot A = 0,$$ the lines are parallel to the surface.
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: