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The electric potential in a region of space is V=(350V.m)/√(x^2+y^2), where x and y are in meters.

  • Calculate the Electric Field strength at (x,y)=(3.0m,\ 1.0m).
  • Find the angle in counterclockwise CCW direction from positive x-axis in which the Electric Field acts at (x,y)=(3.0m,\ 1.0m).
  • Calculate your answer using two significant figures.

The aim of this question is to find the strength of the electric field at the given coordinates created by the given electric potential, its direction at the given coordinates, and its angle with reference to positive x-axis.

The basic concept behind this article is the Electric Potential. It is defined as the total potential which causes a unit electrical charge to move between two points in an electric field. The Electric Field of Potential V can be calculated as follows:

\[E=-\vec{\nabla}V=-(\frac{\partial\ V}{\partial\ x}\hat{i}+\frac{\partial\ V}{\partial\ y}\hat{j})\]

Expert Answer

Given Electric potential:

\[V\ =\ \frac{350\ V.\ m}{\sqrt{x^2+y^2}}\]

Electric field:

\[\vec{E}=-\vec{\mathrm{\nabla}}\ V\]

\[\vec{E}=- \left(\hat{i}\frac{\partial V}{\partial x}+\hat{j}\frac{\partial V}{\partial y}\right)\]

Now putting equation of $V$ here:

\[\vec{E}=- \left(\hat{i}\frac{\partial}{\partial x}\left[\frac{350\ V.\ m}{\sqrt{x^2+y^2}}\right]+\hat{j}\frac{\partial V}{\partial y}\ \left[\frac{350\ V.\ m}{\sqrt{x^2+y^2}}\right]\right)\]

Taking derivative:

\[\vec{E}=-(350\ V.\ m)\ \left(\hat{i}\frac{\partial}{\partial x}\left[\frac{1}{\sqrt{x^2+y^2}}\right]+\hat{j}\frac{\partial V}{\partial y}\ \left[\frac{1}{\sqrt{x^2+y^2}}\right]\right)\]

\[\vec{E}=-(350\ V.\ m)\ \left(\hat{i}\left[\frac{-1}{2}\ {(x^2+y^2)}^\frac{-3}{2}\ (2x+0)\right]+\hat{j}\ \left[\frac{-1}{2}\ {(x^2+y^2)}^\frac{-3}{2}\ (0+2y)\right]\right)\]

\[\vec{E}=-(350\ V.\ m)\ \left(\hat{i}\left[\frac{-x}{{(x^2+y^2)}^\frac{3}{ 2}}\right]+\hat{j}\ \left[\frac{-y}{{(x^2+y^2)}^\frac{3}{2}}\right]\right)\]

\[\vec{E}=\hat{i}\left[\frac{\left(350\ V.\ m\right)x}{ \left(x^2+y^2\right)^\frac{3}{2}}\right]+\hat{j}\ \left[\frac{\left(350\ V.\ m\right)y}{ \left(x^2+y^2\right)^\frac{3}{2 }}\right]\]

The Electric field at $(x, y) = (3 m, 1 m)$ is:

\[\vec{E}= \hat{i}\left[ \frac{\left(350\ V.\ m\right)(3)}{\left(3^2+1^2\right)^\frac{3}{2}}\right]+\hat{j}\ \left[\frac{\left(350\ V.\ m\right)(1)}{\left(3^2+1^2\right)^\frac{3}{2}}\right]\]

\[\vec{E}=33.20\ \hat{i}+11.07\ \hat{j}\ \]

Strength of Electric field at $(x, y) = (3 m, 1m)$ will be:

\[\vec{E}=\sqrt{\left(33.20\right)^2\ \hat{i}+\left(11.07\right)^2\ \hat{j}}\]

\[\vec{E}=\sqrt{ 1224.78}\]

\[\vec{E} =35.00\]

The Direction of Electric field at $(x, y) = (3 m, 1m)$ will be:

\[\theta\ =\ \tan^{-1}{\frac{11.07}{33.20}}\]

\[\theta\ =\ 18.44°\]

Numerical Results

Strength of Electric field at $(x, y) = (3 m, 1m)$ is:

\[\vec{E}=\sqrt{\left(33.20\right)^2\ \hat{i}+\left(11.07\right)^2\ \hat{j}}\]

\[\vec{E} =35.00\]

The Direction of Electric field at $(x, y) = (3 m, 1m)$ is:

\[\theta\ =\ 18.44°\]

Example

The electric potential in a region of space is $V = \frac{250\ V.\ m}{\sqrt{x^2+y^2}}$. Calculate the Electric Field strength and the angle in counterclockwise $CCW$ direction from positive $x-axis$ at $(x,y)=(3.0m,\ 1.0m)$.

Given Electric potential:

\[V\ =\ \frac{250\ V.\ m}{\sqrt{x^2+y^2}}\]

Electric field:

\[\vec{E}=-\vec{\mathrm{\nabla}}\ V\]

\[\vec{E}=- \left(\hat{i}\frac{\partial V}{\partial x}+\hat{j}\frac{\partial V}{\partial y}\right)\]

Now putting equation of $V$ here:

\[\vec{E} = – \left(\hat{i}\frac{ \partial}{ \partial x}\left[ \frac{250\ V.\ m}{ \sqrt{x^2+y^2}}\right]+\hat{j}\frac{ \partial V}{ \partial y}\ \left[ \frac{250\ V.\ m}{\sqrt{x^2+y^2}} \right] \right)\]

Taking derivative:

\[\vec{E} = -( 250\ V.\ m)\ \left(\hat{i}\frac{\partial}{ \partial x}\left[ \frac{1}{\sqrt{x^2+y^2}}\right]+\hat{j}\frac{ \partial V}{ \partial y}\ \left[ \frac{1}{\sqrt{x^2+y^2}}\right]\right)\]

\[\vec{E} =-(250\ V.\ m)\ \left(\hat{i}\left[\frac{-1}{2}\ {(x^2+y^2)}^\frac{-3}{ 2}\ (2x+0)\right]+\hat{j}\ \left[ \frac{-1}{2}\ {(x^2+y^2)}^\frac{-3}{ 2}\ (0+2y) \right]\right)\]

\[\vec{E} =-(250\ V.\ m)\ \left(\hat{i}\left[ \frac{-x}{{(x^2+y^2)}^\frac{3 }{2}} \right]+\hat{j}\ \left[ \frac{-y}{{(x^2+y^2)}^\frac{ 3}{2}} \right]\right)\]

\[\vec{E} =\hat{i}\left[\frac{ \left(250\ V.\ m\right)x}{\left(x^2+y^2\right)^\frac{3}{2}} \right]+\hat{j}\ \left[\frac{ \left(250\ V.\ m\right)y}{\left(x^2+y^2\right)^\frac{3}{2}} \right]\]

The Electric field at $(x, y) = (3 m, 1 m)$ is:

\[\vec{E}= \hat{i} \left[ \frac{\left(250\ V.\ m\right)(3)}{ \left(3^2+1^2\right)^\frac{ 3}{2}} \right]+\hat{ j}\ \left[ \frac{\left(250\ V.\ m\right)(1)}{ \left(3^2+1^2\right)^\frac{3 }{ 2}} \right]\]

\[\vec{E}=23.72\ \hat{i}+7.90\ \hat{j}\ \]

Strength of Electric field at $(x, y) = (3 m, 1m)$ will be:

\[\vec{E} =\sqrt{ \left(23.72 \right)^2\ \hat{i}+\left(7.90\right)^2\ \hat{j} }\]

\[\vec{E}=\sqrt{ 625.05}\]

\[\vec{E} =25.00\]

The Direction of Electric field at $(x, y) = (3 m, 1m)$ will be:

\[\theta\ =\ \tan^{-1}{\frac{7.90}{23.72}}\]

\[\theta\ =\ 18.42°\

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