Electric potential at the vertex of a triangle

In summary, the potential at the vertex P can be written as $$\phi_P= (σb/4π \varepsilon_o)ln[(1+sinθ)/cosθ]$$
  • #1
math4everyone
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0
Member advised to use the homework template for posts in the homework sections of PF.
I am stuck with this problem:

The right triangle shown with vertex P at the origin has base b, altitude a, and uniform density of surface charge σ. Determine the potential at the vertex P. First find the contribution of the vertical strip of width dx at x. Show that the potential at P can be written as $$\phi_P= (σb/4π \varepsilon_o)ln[(1+sinθ)/cosθ]$$
MaW6n.png

What I have done so far:

A tiny box of the strip has width dx and height $$rd\theta$$ so its contribution to P is
\begin{equation}
d\phi = \frac{(\sigma)(dx)(rd\theta)}{4 \pi \varepsilon_o r}=\frac{(\sigma)(dx)(d\theta)}{4 \pi \varepsilon_o }
\end{equation}
So the contribution of the entire strip would be
\begin{equation}
\frac{\sigma dx}{4 \pi \varepsilon_o}\int_0^{arctan(\frac{a}{b})} d\theta = \frac{\sigma dx}{4 \pi \varepsilon_o} arctan(\frac{a}{b})
\end{equation}
So the potential, $$\phi$$ would be:
\begin{equation}
\phi = \frac{\sigma}{4 \pi \varepsilon_o} arctan(\frac{a}{b}) \int_0^b dx = b\frac{\sigma}{4 \pi \varepsilon_o} arctan(\frac{a}{b})
\end{equation}
But this is clearly not the answer. What am I doing wrong?
 
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  • #2
Look at the drawing. Angle θ is fixed, so dθ has no meaning here. The height of the strip is better represented as y and the area element as dA = y dx. Then express y in terms of x (it's a straight line) and integrate. Of course, first you need to find the potential of a line of charge of length y at perpendicular distance x from one of its ends.
 
  • #3
math4everyone said:
A tiny box of the strip has width dx and height $$rd\theta$$
As kuruman pointed out, ##\theta## is a fixed angle.

Maybe you were thinking of using a variable angle ##\varphi##, say, as shown in the attached figure below.
But, note that the height, ##dy##, of the small box is not equal to ##r d\varphi##.

Can you see how to express ##dy## in terms of ##r d\varphi## and an appropriate trig function of ##\varphi##? If so, then you can let the integration variable for the vertical strip be ##\varphi## and it should simplify nicely.
 

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  • #4
You probably could see the area of your element as dydx and integrate your potential function over the entire surface (one tip: try using polar coordinates).
 

1. What is the formula for calculating the electric potential at the vertex of a triangle?

The formula for calculating the electric potential at the vertex of a triangle is V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge, and r is the distance from the charge to the vertex.

2. How is the electric potential affected by the distance from the charge to the vertex?

The electric potential at the vertex of a triangle is inversely proportional to the distance from the charge to the vertex. This means that as the distance increases, the electric potential decreases.

3. What does the sign of the charge tell us about the electric potential at the vertex of a triangle?

The sign of the charge determines the direction of the electric potential at the vertex of a triangle. A positive charge will have a positive electric potential, while a negative charge will have a negative electric potential.

4. Can the electric potential at the vertex of a triangle be negative?

Yes, the electric potential at the vertex of a triangle can be negative. This indicates that the direction of the electric field points towards the charge rather than away from it.

5. How does the number of charges in a triangle affect the electric potential at the vertex?

The electric potential at the vertex of a triangle is the sum of the electric potentials from each charge. Therefore, the more charges in the triangle, the higher the electric potential at the vertex will be.

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