Electric field vector in a region is given by $\mathbf{E}=(3 \hat{\mathbf{i}}+4 y \hat{\mathbf{j}})…

Electric field vector in a region is given by $\mathbf{E}=(3 \hat{\mathbf{i}}+4 y \hat{\mathbf{j}}) \mathrm{V}-\mathrm{m}^{-1}$. The potential at the origin is zero. Then, the potential at a point $(2,1) \mathrm{m}$ is
  1. 7 V
  2. 8 V
  3. -8 V
  4. -7 V

Solution

$ \begin{gathered} \text { As, } \quad E=\frac{-\partial V_x}{\partial x} \hat{\mathrm{i}}+\frac{-\partial V_y}{\partial y} \hat{\mathrm{j}} \\ \frac{-\partial V_x}{\partial x}=3 \text { and } \frac{-\partial V_y}{\partial y}=4 y \\ \Rightarrow \quad V_x=-3 x \text { and } V_y=-2 y^2 \end{gathered} $ So, $V=$ potential function $ =-\left(3 x+2 y^2\right) $ Potential at point $(2,1)$ is $ V=-\left(3 \times 2+2 \times 1^2\right)=-8 \mathrm{~V} $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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