Question

Which of the following statement(s) is/are correct.
This question has multiple correct options.
A) If the electric field due to a point charge varies as ${{r}^{-25}}$ instead of ${{r}^{-2}}$, then the Gauss law will still be valid.
B) The gauss law can be used to calculate the field distribution around an electric dipole.
C) If the electric field between two-point charges is zero somewhere, then the sign of the two charges is the same.
D) The work done by the external force in moving unit positive charge from point $A$ at potential ${{V}_{A}}$ to point $B$ at potential ${{V}_{B}}$ is $({{V}_{B}}-{{V}_{A}})$.

Answer 1

By applying Gauss law in options (A) and (B), we can easily check that correctness of both these statements. With the expression for the electric field joining two points, we may check if option (C) is correct or wrong. From the definition of potential difference, we also can check if option (D) is correct or wrong.

Formula used:
$1)E=k{{r}^{-25}}$
$2){{\Phi }_{E}}=\oint{E.dA=}\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{o}}}$
$3)W=q({{V}_{A}}-{{V}_{B}})$

Complete answer:
We know that Gauss law gives an expression for the electric field flux through a gaussian surface, proportional to the electric charge enclosed inside this surface. Mathematically, Gauss law is given by
${{\Phi }_{E}}=\oint{E.dA=}\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{o}}}$
where
${{\Phi }_{E}}$ is the electric field flux through a gaussian surface
$E$ is the electric field strength through this surface
$dA$ is the area of this surface
${{Q}_{enclosed}}$ is the amount of electric charge enclosed within this surface
${{\varepsilon }_{0}}$ is the electric constant
Let this be equation 1.
Now, if electric field due to point charge varies as ${{r}^{-25}}$ instead of ${{r}^{-2}}$, then, we can express $E$as:
$E=k{{r}^{-25}}$
where
$k$ is a constant.
Let this be equation 2.
Substituting equation 2 and $dA=4\pi {{r}^{2}}$ in equation 1, we have
$\oint{E.dA=k{{r}^{-25}}}\times 4\pi {{r}^{2}}=4\pi k{{r}^{-23}}\ne \dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{o}}}$
Therefore, we can conclude that Gauss law is not valid in such a case and hence, option A is incorrect.
Moving on to the next given option, we already know that charges on both the ends of a dipole are opposite and thus, the total enclosed charge will sum up to zero$({{Q}_{enclosed}}=e-e=0)$, if a gaussian surface is considered on the same. Therefore, we can conclude that option B is also incorrect.
Now, electric field $({{E}_{Q}})$ in the line connecting two point charges ${{Q}_{1}}$ and ${{Q}_{2}}$ is given by
${{E}_{Q}}={{E}_{{{Q}_{1}}}}+{{E}_{{{Q}_{2}}}}$
where
${{E}_{{{Q}_{1}}}}$ is the electric field due to ${{Q}_{1}}$
${{E}_{{{Q}_{2}}}}$ is the electric field due to ${{Q}_{2}}$
Let this be equation 3.
We also know that electric field $(E)$ due to a positive charge $(q)$ at a distance $r$ is given by
$E=\dfrac{kq}{{{r}^{2}}}$
where
$k=const$
Let this be equation 3.
From option C, it is given that
${{E}_{Q}}=0$
Let this be condition X.
Using equation 3, for the condition X to be true, we have
${{E}_{Q}}=0\Rightarrow {{E}_{{{Q}_{1}}}}+{{E}_{{{Q}_{2}}}}=0\Rightarrow {{E}_{{{Q}_{1}}}}={{E}_{{{Q}_{2}}}}$
Let this be equation 4.
From equation 4, it is clear that both electric fields can be due to similar charges, acting in directions opposite to each other.
Therefore, we can conclude that option C is correct.
Moving on to the next option, we already know that work done by external force to move a positive charge from point $A$ to another point $B$ is given by
$W=q({{V}_{A}}-{{V}_{B}})$
where
$W$is the work done
$q$ is the charge which is being moved from $A$ to $B$
${{V}_{A}}$ is the potential at point $A$
${{V}_{B}}$ is the potential at point $B$
Let this be equation 5.
Now, if the charge is considered to be a unit charge $(q=1)$, then, equation 5 can be rewritten as
$W=q({{V}_{A}}-{{V}_{B}})=1({{V}_{A}}-{{V}_{B}})=({{V}_{A}}-{{V}_{B}})$
Hence, we can conclude that option D is also correct.

To conclude, options A and B are incorrect while options C and D are correct.

Note:
It is important that we go through the options clearly and carefully, with patience. When proceeded accordingly, we can easily arrive at our final answers easily. For example, if we have understood the concept of Gauss law clearly, we can easily arrive at the correctness of the first two options, without wasting much of our time. Students should also understand that option C is just one of the possibilities with respect to the given condition here.