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Question: If a potential at a point P(1,1,1) is given by the relation \({{V}_{r}}={{x}^{2}}-{{y}^{2}}+2z\) the...

If a potential at a point P(1,1,1) is given by the relation Vr=x2−y2+2z{{V}_{r}}={{x}^{2}}-{{y}^{2}}+2z then calculate the electric field at point P.

Explanation

Solution

In the above question we basically have to determine the electric field for a given variation of potential in that particular region or space. Electric field is nothing but the gradient of the potential. Therefore we will take the gradient of the above relation of potential and then accordingly determine the field at point P.

Formula used:
∂V∂r=∂Vx∂x+∂Vy∂y+∂Vz∂z\dfrac{\partial V}{\partial r}=\dfrac{\partial {{V}_{x}}}{\partial x}+\dfrac{\partial {{V}_{y}}}{\partial y}+\dfrac{\partial {{V}_{z}}}{\partial z}
∇⋅V=∂V∂r=E\nabla \cdot V=\dfrac{\partial V}{\partial r}=E

Complete step by step answer:
Let us say in a given space due to some charge distribution there exists electric potential governed by a particular relation. If V is the electric potential at any point in a given space, then the gradient of the potential at that particular point will give us the magnitude of the electric field(E). Mathematically this can be represented as,
∇⋅V=∂V∂r=E\nabla \cdot V=\dfrac{\partial V}{\partial r}=E
Gradient of the potential is nothing but the dot product of the respective partial derivative in space with respective components in space. If in a particular space, is the potential is given by,
V=Vx+Vy+VzV={{V}_{x}}+{{V}_{y}}+{{V}_{z}} then the gradient of potential i.e. electric field at any point P is given by,
∂V∂r=∂Vx∂x+∂Vy∂y+∂Vz∂z\dfrac{\partial V}{\partial r}=\dfrac{\partial {{V}_{x}}}{\partial x}+\dfrac{\partial {{V}_{y}}}{\partial y}+\dfrac{\partial {{V}_{z}}}{\partial z}
In the above question it is given that the potential varies in space by a relation i.e. Vr=x2−y2+2z{{V}_{r}}={{x}^{2}}-{{y}^{2}}+2z. Therefore by taking the gradient of this function of potential we get,
E=∂V∂r=∂(x2−y2+2z)∂x+∂(x2−y2+2z)∂y+∂(x2−y2+2z)∂z ⇒E=2x−2y+2 \begin{aligned} & E=\dfrac{\partial V}{\partial r}=\dfrac{\partial ({{x}^{2}}-{{y}^{2}}+2z)}{\partial x}+\dfrac{\partial ({{x}^{2}}-{{y}^{2}}+2z)}{\partial y}+\dfrac{\partial ({{x}^{2}}-{{y}^{2}}+2z)}{\partial z} \\\ & \Rightarrow E=2x-2y+2 \\\ \end{aligned}
In the question it is asked to determine the field at point P(1,1,1). Hence from above expression the field at point P is equal to,
E=2x−2y+2 ⇒E=2(1)−2(1)+2 ⇒E=2V/m \begin{aligned} & E=2x-2y+2 \\\ & \Rightarrow E=2(1)-2(1)+2 \\\ & \Rightarrow E=2V/m \\\ \end{aligned}

Therefore the electric field at point P is equal to 2V/m.

Note: The partial derivative of a variable with respect to some other running variable is zero. It is also to be noted that the above electric field is constant only in z direction but varies with x and y. Hence from this we can imply that the field is not uniform in given space.