Question

Electric potential in space is given by V = 9 (x ^3 - y ^3) where x and y are the cartesian coordinate. The electric field vector in appropriate units is
1 -27x^2i + 27y^2j
2 27xi + 27yi
3 27x^2i + 27 xy^2j
4 - 27xyi + 27 xyj

Answer 1

The correct option is, $\bar{E}=-27 x^{2} \i+27 y^{2} \jmath$.

What is electric potential in simple words?

• Electric potential, the amount of work needed to move a unit charge from a reference point to a specific point against an electric field.
• Typically, the reference point is Earth, although any point beyond the influence of the electric field charge can be used.

According to the question:

$E=E_{x}+E_{y}\\E\vec{E}_{x}=\frac{-\partial V}{\partial x} \hat{\imath}=\frac{-\partial\left(9 x^{3}-y^{3}\right)}{\partial x}=-27 x^{2} \hat{\imath}\\E_{y}=-\frac{\partial V}{\partial y} \hat{\jmath}=\frac{-\partial\left(9 x^{3}-y^{3}\right)}{\partial y}=27 y^{2} \hat{\jmath}$

$\bar{E}=-27 x^{2} \i+27 y^{2} \jmath$

• Partial differentiation of V w.r.t. x means we differentiate V w.r.t. x assuming y and z constant.
• Partial differentiation of V w.r.t. y means we differentiate V w.r.t. y assuming x and z constant.
• Partial differentiation of V w.r.t. z means we differentiate V w.r.t. z assuming x and y constant.

Hence,  The electric field vector in appropriate units is, $\bar{E}=-27 x^{2} \i+27 y^{2} \jmath$.

#SPJ2