Physics doubt!

If \(\vec{E}\) is given by\(\vec{E} = y \hat{i} +x \hat{j}\) , then find the the potential \(V_{(x,y)}\) if \(V_{∞}\) is 10 \(V\).

I did

∫10V(x,y)dV=−∫∞(x,y)(yi^+xj^).(dxi^+dyj^) \int_{10}^{V_{(x,y)}} dV= -\int_{∞}^{(x,y)} (y \hat{i} +x \hat{j}).(dx \hat{i} +dy \hat{j})

Substitute z=xyz=xy

V(x,y)−10=[z]∞xyV_{(x,y)} - 10 = [z]^{xy} _ ∞

V(x,y)−10=xy−∞\color{#D61F06}{V_{(x,y)} - 10 = xy - ∞}

Why that happened

[Edit] - Issue resolved , thanks soumo for help.

Note by Kushal Patankar
6 years, 1 month ago

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1 vote

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Comments

Here Electric field at infity is infinite , so you can't take potential reference as 10 volt at that place (region) . Since E.F here has trajectory of hyperbola , which is unbound at very very very large distance from origin , So potential at infinite being 10 volt is meaningless .

Nishu sharma - 6 years, 1 month ago

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I didn't got why or how it is meaningless

Kushal Patankar - 6 years, 1 month ago

  • Did you create this problem?

  • I want you to spot a fallacy (if it's there) in the following :

V(A)−V(B)=−∫BAE.drV(A)−V(∞)=−∫∞AE.drV(A)=−E.A+E.∞ { V }_{ \left( A \right) }-{ V }_{ \left( B \right) }=-\int _{ B }^{ A }{ E.dr } \\ { V }_{ \left( A \right) }-{ V }_{ \left( \infty \right) }=-\int _{ \infty }^{ A }{ E.dr } \\ { V }_{ \left( A \right) }=-E.A+E.\infty

I have assumed V∞=0\displaystyle { V }_{ \infty }=0 and EE to be constant.

So, what is wrong? Is even anything wrong?

  • We know that V(A)−V(B)=−∫BAE.dr\displaystyle { V }_{ \left( A \right) }-{ V }_{ \left( B \right) }=-\int _{ B }^{ A }{ E.dr } . To find the electric potential at AA we choose BB to be a reference point. It is very nice if we choose BB such that the electric potential at BB is zero. According to your EE (electric field), potential is zero at x=0,y=0x=0,y=0 i.e., at the origin.

  • So do we need to find V(A)−V(∞)\displaystyle { V }_{ \left( A \right) }-{ V }_{ \left( \infty \right) }? or V(A)−V(0)\displaystyle { V }_{ \left( A \right) }-{ V }_{ \left( 0 \right) } i.e. V(A)\displaystyle { V }_{ \left( A \right) }

Only after making some assumptions, I came across the following approach:

If V(A)−V(B)=−∫BAE.dr\displaystyle { V }_{ \left( A \right) }-{ V }_{ \left( B \right) }=-\int _{ B }^{ A }{ E.dr }

Then V(A)−V(0)=−∫0AE.dr\displaystyle { V }_{ \left( A \right) }-{ V }_{ \left( 0 \right) }=-\int _{ 0 }^{ A }{ E.dr } gives the potential at AA w.r.t origin 0,00,0.

On the same basis V∞−V(0)=−∫0AE.dr{ V }_{ \infty }-{ V }_{ \left( 0 \right) }=-\int _{ 0 }^{ A }{ E.dr } gives the potential at infinity w.r.t to origin.

And (V(A)−V(0))−(V∞−V(0))\displaystyle \left( { V }_{ \left( A \right) }-{ V }_{ \left( 0 \right) } \right) -\left( { V }_{ \infty }-{ V }_{ \left( 0 \right) } \right) gives potential at AA w.r.t infinity.

  • Now let me return to the very first question: 'Did you create this problem?'. If yes, then I think you can polish it by providing a relation between xx & yy. I am saying so because only after trying to find the role of xx & yy, I stumbled upon the idea of trying to have x=0,y=0x=0,y=0.

Hope this helps you to find the answer(s) even if this isn't the answer.

:)

Soumo Mukherjee - 6 years, 1 month ago

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The problem is that, if I consider any point in x-y plane I get answer of electric potential as -infinity .I am unable to figure out what's the problem.

Why is relation between x and y required?

Kushal Patankar - 6 years, 1 month ago

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The ultimate aim is to find the potential at a point w.r.t another point where we can reasonably take the potential to be zero. This reference point need not necessarily be infinity.

I wanted you to spot the fallacy in second point of my previous comment. That way we always get potential at any point to be infinity. That approach is wrong.

Did you try this (V(A)−V(0))−(V∞−V(0))\displaystyle \left( { V }_{ \left( A \right) }-{ V }_{ \left( 0 \right) } \right) -\left( { V }_{ \infty }-{ V }_{ \left( 0 \right) } \right) ?

Soumo Mukherjee - 6 years ago
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