A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be
A.
due to $${q_1}$$ and $${q_2}$$ only
B.
due to $${q_2}$$ only
C.
zero
D.
due to all
Answer :
due to all
Solution :
The flux through the Gaussian surface is due to the charges inside the Gaussian surface. But the electric field on the Gaussian surface will be due to the charges present in side the Gaussian surface and outside it. It will be due to all the charges.
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: