Waste less time on Facebook — follow Brilliant.
×

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

\[ \int_{10}^{V_{(x,y)}} dV= -\int_{∞}^{(x,y)} (y \hat{i} +x \hat{j}).(dx \hat{i} +dy \hat{j})\]

Substitute \(z=xy\)

\[V_{(x,y)} - 10 = [z]^{xy} _ ∞\]

\[\color{red}{V_{(x,y)} - 10 = xy - ∞}\]

Why that happened

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

Note by Kushal Patankar
1 year, 10 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

  • Did you create this problem?

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

\[ { V }_{ \left( A \right) }-{ V }_{ \left( B \right) }=-\int _{ B }^{ A }{ E.dr } \\ { V }_{ \left( A \right) }-{ V }_{ \left( \infty \right) }=-\int _{ \infty }^{ A }{ E.dr } \\ { V }_{ \left( A \right) }=-E.A+E.\infty \]

I have assumed \(\displaystyle { V }_{ \infty }=0\) and \(E\) to be constant.

So, what is wrong? Is even anything wrong?

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

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

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

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

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

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

And \(\displaystyle \left( { V }_{ \left( A \right) }-{ V }_{ \left( 0 \right) } \right) -\left( { V }_{ \infty }-{ V }_{ \left( 0 \right) } \right) \) gives potential at \(A\) 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 \(x\) & \(y\). I am saying so because only after trying to find the role of \(x\) & \(y\), I stumbled upon the idea of trying to have \(x=0,y=0\).

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

:) Soumo Mukherjee · 1 year, 10 months ago

Log in to reply

@Soumo Mukherjee 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 · 1 year, 10 months ago

Log in to reply

@Kushal Patankar 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 \(\displaystyle \left( { V }_{ \left( A \right) }-{ V }_{ \left( 0 \right) } \right) -\left( { V }_{ \infty }-{ V }_{ \left( 0 \right) } \right) \)? Soumo Mukherjee · 1 year, 10 months ago

Log in to reply

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 · 1 year, 10 months ago

Log in to reply

@Nishu Sharma I didn't got why or how it is meaningless Kushal Patankar · 1 year, 10 months ago

Log in to reply

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...