It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss's theorem because

In a region,the electric field is given by $\overrightarrow{E}=(3 \hat{i}+5 \hat{j}+7 \hat{k}) \text{ NC}^{-1}$. The electric flux through a surface of area $3 \text{ m}^2$ in the $yz$-plane is (in $SI$ units):

$A$ charge $Q$ is enclosed by a Gaussian surface of radius $R$. If the radius is doubled,then the outward electric flux will

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:

Select the correct statement$(s)$.
$(1)$ The tangent drawn at any point on an electric field line gives the direction of the force acting on a positive charge at that point.
$(2)$ The normal drawn at any point on an electric field line gives the direction of the force acting on a positive charge at that point.
$(3)$ Electric field lines start from a negative charge and end on a positive charge.
$(4)$ Electric field lines start from a positive charge and end on a negative charge.

An electrostatic field line leaves at an angle $\alpha$ from a point charge $q_{1}$ and connects with a point charge $-q_{2}$ at an angle $\beta$ ($q_{1}$ and $q_{2}$ are positive). See the figure below. If $q_{2} = \frac{3}{2} q_{1}$ and $\alpha = 30^{\circ}$,then: