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

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

Explanation

Solution

The relation between the potential and the electric field at a point in space is expressed as the gradient of the other. Using this we will first obtain the general relation between the electric and the potential for the electric potential given in the question. Hence then substituting the coordinate of point we will obtain the potential at any point in space.
Formula used:
E→=−∂V∂r\overrightarrow{E}=-\dfrac{\partial V}{\partial r}

Complete step by step answer:
Let us say there exists an electric potential (V) in any given space. The electric field (E) at any point (r) in space is given by the negative gradient of potential. Mathematically expressed as,
E→=−∂V∂r\overrightarrow{E}=-\dfrac{\partial V}{\partial r}
The above equation in terms of the coordinate system is given by,
−∂V∂r=−[∂V∂xi^+∂V∂yj^+∂V∂zk^]-\dfrac{\partial V}{\partial r}=-\left[ \dfrac{\partial V}{\partial x}\widehat{i}+\dfrac{\partial V}{\partial y}\widehat{j}+\dfrac{\partial V}{\partial z}\widehat{k} \right]
The potential at any point in the space is given as,
Vp=x2+y2+2{{V}_{p}}={{x}^{2}}+{{y}^{2}}+2 . Therefore from the above equation the electric field at a point is equal to,
E→=−∂Vp∂r=−[∂Vp∂xi^+∂Vp∂yj^+∂Vp∂zk^] E→=−[∂(x2+y2+2)∂xi^+∂(x2+y2+2)∂yj^+∂(x2+y2+2)∂zk^] ⇒E→=−[2xi^+2yj^+0k^] \begin{aligned} & \overrightarrow{E}=-\dfrac{\partial {{V}_{p}}}{\partial r}=-\left[ \dfrac{\partial {{V}_{p}}}{\partial x}\widehat{i}+\dfrac{\partial {{V}_{p}}}{\partial y}\widehat{j}+\dfrac{\partial {{V}_{p}}}{\partial z}\widehat{k} \right] \\\ & \overrightarrow{E}=-\left[ \dfrac{\partial ({{x}^{2}}+{{y}^{2}}+2)}{\partial x}\widehat{i}+\dfrac{\partial ({{x}^{2}}+{{y}^{2}}+2)}{\partial y}\widehat{j}+\dfrac{\partial ({{x}^{2}}+{{y}^{2}}+2)}{\partial z}\widehat{k} \right] \\\ & \Rightarrow \overrightarrow{E}=-\left[ 2x\widehat{i}+2y\widehat{j}+0\widehat{k} \right] \\\ \end{aligned}
Therefore the electric field at point P (1,1,1) is equal to,
E→=−[2xi^+2yj^+0k^] E→=−[2i^+2j^]V/m \begin{aligned} & \overrightarrow{E}=-\left[ 2x\widehat{i}+2y\widehat{j}+0\widehat{k} \right] \\\ & \overrightarrow{E}=-\left[ 2\widehat{i}+2\widehat{j} \right]V/m \\\ \end{aligned}
More precisely speaking the magnitude of the electric field at point P is equal to,
∣E→∣=22+22V/m ∣E→∣=22V/m \begin{aligned} & \left| \overrightarrow{E} \right|=\sqrt{{{2}^{2}}+{{2}^{2}}}V/m \\\ & \left| \overrightarrow{E} \right|=2\sqrt{2}V/m \\\ \end{aligned}

Therefore the electric field at a point P is equal to E→=−[2i^+2j^]V/m\overrightarrow{E}=-\left[ 2\widehat{i}+2\widehat{j} \right]V/m whose magnitude is equal to 22V/m2\sqrt{2}V/m.

Note: It is to be noted that the potential is a scalar quantity. When we operate the gradient operator on the potential we get an electric field which is a vector quantity. This is because the gradient is a directional derivative and hence the quantity always has a direction associated with it. The derivative of quantity with the same variable is a finite whereas the derivative with respect to some other variable of a quantity is zero.