Question

Determine the electric field strength vector if the potential of this field depends on $x$,$y$ coordinate as $V=10axy$
\[\begin{align}
  & A.10a(y\hat{i}+x\hat{j}) \\
 & B.-10a(y\hat{i}+x\hat{j}) \\
 & C.-a(y\hat{i}+x\hat{j}) \\
 & D.-10a(x\hat{i}+y\hat{k}) \\
\end{align}\]

Answer 1

Hint: Electric field is the electric force due to a unit positive charge which is at rest would exert on its surrounding. We also know that the electric potential due to a charge, is defined as the amount of energy needed to move a unit positive charge to infinity. Using the relation between the two we can solve this sum.

Formula used: $E=\dfrac{V}{r}$ and $E=-\nabla V$

Complete step by step answer:
We know that the electric force due to a pair of charges is given by Coulomb's law. An electric field can be produced by a time-varying electric field or an electrical charge. These can be either attracting or repelling in nature.
An electric field E is defined as the electric force F per unit positive charge q , which is infinitesimally small and at rest, and is given as $E=\dfrac{F}{q}$. Then$E=\dfrac{kq}{r^{2}}$, where $k=\dfrac{1}{4\pi\epsilon_{0}}$ which is a constant and $r$ is the distance between the unit charges. Since the electric field is a vector quantity, it acts along the direction of the distance $r$ , then we can denote it as $\vec E=\dfrac{kq\vec r}{r^{3}}$.
We also know that the electric potential due to a charge is defined as the amount of energy needed to move a unit positive charge to infinity. Also the potential at any point is the vector sum of potentials at that point. Also, potential is proportional to the charge and inversely proportional to the square of the distance between the point and the charge. $V=\dfrac{kq}{r}$
Then clearly, we can say that, $E=\dfrac{V}{r}$ where $r$ is the distance of the charge from the unit charge.
Also, $E=-\nabla V$, where, $\nabla V$ is called the potential gradient
 $\implies E=-\left(\dfrac{dV}{dx}\hat i+\dfrac{dV}{dy}\hat j\right)$
Given that $V=10axy$ then substituting the values, we have,
$\implies E=-\left(\dfrac{10axy}{dx}\hat i+\dfrac{10axy}{dy}\hat j\right)$
$\implies E=-\left(10ay \hat i+10ax\hat j\right)$
$\implies E=-10a\left(y \hat i+x\hat j\right)$

So, the correct answer is “Option B”.

Note: Electric field is in the direction of the force. Usually, the electric field of a point positive charge is radially outwards, whereas the electric field of a point negative charge is radially inwards to the charge. However, the electric field also depends on the symmetry of the charge carrying conductor.