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BioPhysicist
BioPhysicist
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Read Edit: You are right! The definition of potential given by $$V(\mathbf{r})\equiv -\int_{\mathcal{O}}^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$$ In your case the $\mathbf{E}$ is outward in x-direction while the displacement vector $d\mathbf{l}=\mathbf{x}-(\mathbf{x}+d\mathbf{x})=-d\mathbf{x}$. So yes the dot product between them will be negative.}

The following possibility may occur because they didn't give the value of $k$ it might be positive or negative, so may be they have taken it to be negative. Apart from this your reasoning is correct.

Edit : Actually there is flaw in this thinking. While we derive the expression for line intergral we don't care for the sign of $\Delta x$ because that already taken into account with limits of integration.For example : consider the example given below in the comment box.

So it's as given in the book.

Read Edit: You are right! The definition of potential given by $$V(\mathbf{r})\equiv -\int_{\mathcal{O}}^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$$ In your case the $\mathbf{E}$ is outward in x-direction while the displacement vector $d\mathbf{l}=\mathbf{x}-(\mathbf{x}+d\mathbf{x})=-d\mathbf{x}$. So yes the dot product between them will be negative.}

The following possibility may occur because they didn't give the value of $k$ it might be positive or negative, so may be they have taken it to be negative. Apart from this your reasoning is correct.

Edit : Actually there is flaw in this thinking. While we derive the expression for line intergral we don't care for the sign of $\Delta x$ because that already taken into account with limits of integration.For example : consider the example given below in the comment box.

So it's as given in the book.

While we derive the expression for line intergral we don't care for the sign of $\Delta x$ because that already taken into account with limits of integration.For example : consider the example given below in the comment box.

So it's as given in the book.

Answer corrected.
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Himanshu
Himanshu
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You are right! The definition of potential given by $$V(\mathbf{r})\equiv -\int_{\mathcal{O}}^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$$Read Edit: In your case the You are right! The definition of potential given by $$V(\mathbf{r})\equiv -\int_{\mathcal{O}}^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$$ In your case the $\mathbf{E}$ is outward in x-direction while the displacement vector $d\mathbf{l}=\mathbf{x}-(\mathbf{x}+d\mathbf{x})=-d\mathbf{x}$. So yes the dot product between them will be negative.}

The following possibility may occur because they didn't give the value of $k$ it might be positive or negative, so may be they have taken it to be negative. Apart from this your reasoning is correct.

Edit $\mathbf{E}$: Actually there is outwardflaw in x-direction whilethis thinking. While we derive the displacement vectorexpression for line intergral we don't care for the sign of $d\mathbf{l}=\mathbf{x}-(\mathbf{x}+d\mathbf{x})=-d\mathbf{x}$$\Delta x$ because that already taken into account with limits of integration. So yesFor example : consider the dot product between them will be negativeexample given below in the comment box.

 

The following possibility may occur because they didn't giveSo it's as given in the value of $k$ it might be positive or negative, so may be they have taken it to be negative. Apart from this your reasoning is correctbook.

You are right! The definition of potential given by $$V(\mathbf{r})\equiv -\int_{\mathcal{O}}^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$$ In your case the $\mathbf{E}$ is outward in x-direction while the displacement vector $d\mathbf{l}=\mathbf{x}-(\mathbf{x}+d\mathbf{x})=-d\mathbf{x}$. So yes the dot product between them will be negative.

 

The following possibility may occur because they didn't give the value of $k$ it might be positive or negative, so may be they have taken it to be negative. Apart from this your reasoning is correct.

Read Edit: You are right! The definition of potential given by $$V(\mathbf{r})\equiv -\int_{\mathcal{O}}^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$$ In your case the $\mathbf{E}$ is outward in x-direction while the displacement vector $d\mathbf{l}=\mathbf{x}-(\mathbf{x}+d\mathbf{x})=-d\mathbf{x}$. So yes the dot product between them will be negative.}

The following possibility may occur because they didn't give the value of $k$ it might be positive or negative, so may be they have taken it to be negative. Apart from this your reasoning is correct.

Edit : Actually there is flaw in this thinking. While we derive the expression for line intergral we don't care for the sign of $\Delta x$ because that already taken into account with limits of integration.For example : consider the example given below in the comment box.

So it's as given in the book.

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Himanshu
Himanshu
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You are right! The definition of potential given by $$V(\mathbf{r})\equiv -\int_{\mathcal{O}}^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$$ In your case the $\mathbf{E}$ is outward in x-direction while the displacement vector $d\mathbf{l}=\mathbf{x}-(\mathbf{x}+d\mathbf{x})=-d\mathbf{x}$. So yes the dot product between them will be negative.

The following possibility may occur because they didn't give the value of $k$ it might be positive or negative, so may be they have taken it to be negative. Apart from this your reasoning is correct.