Answer / Explanation:
For the purpose of clarity, it is to be noted that this question is incomplete due to the fact that the diagrammatic illustration accompanying the question have not been proved.
However, we can find the diagram below.
To be able to properly calculate the electric potential at B, we need to first of all calculate the units if alpha ( ∝)
Therefore:
identifying the units of ∝, we have
λ = c / m = α x = c / m² . m
Therefore, so α has units of c / m².
Now going back to calculate the electric potential at B, according the diagram below that completes the question, we assume that V = 0 at x = infinity, therefore, we treat V as the sum of the point charge potentials from each bit of charge dq.
Therefore,
V = ∫ K dq/r = ∫ lin(l)(o) K λdx/x + d = ∫ lin(l)(o) K ∝xdx/x + d
Now breaking the above down further,
we have,
V = K∝ ∫ lin(l)(o) (x + d - d) dx / x + d = K∝ ∫ lin(l)(o) (1) dx - d.dx / x + d
= K∝ [ ( L - 0) - d ( ln [x + d] ║(ln(L)(0)] = K∝ [ L - d (ln [ L + d] - ln [ d] )]
Therefore,
V = K∝ [ L - d in ( L + d / d).
Therefore, electric potential at point B = V = K∝ [ L - d in ( L + d / d).
The electric potential at point B is in terms of α, ke, L, b, and d is:
[tex]\mathbf{V \Big (\dfrac{L}{2}, b \Big ) = k_e \alpha \dfrac{L}{2} \ In \Bigg ( \dfrac{\sqrt{(\dfrac{L}{2})^2+b^2}+ \dfrac{L}{2}} {\sqrt{(\dfrac{L}{2})^2+b^2}- \dfrac{L}{2}} \Bigg)}[/tex]
The electric potential expresses the level of potential energy that a single point charge has at a specific location. The electric potential of a point is equivalent and proportional to the electric potential energy of each charged particle there divided by the particle's charge in coulombs.
N:B The missing image is attached in the image below.
From the image, let's take the potential at a point p which has coordinates x and y.
Then, the differential integration dV(X, Y) due to a segment dx of the charged rod can be expressed as;
[tex]\mathbf{dV = \dfrac{k_e dx'dx'}{\sqrt{(x'-x)^2+y^2}}}[/tex]
[tex]\mathbf{V = k_e \alpha \int ^L_0 \dfrac{x'dx'}{\sqrt{(x'-x)^2+y^2}}}[/tex]
Suppose we make an assumption that we replace [tex]\mathbf{(x' -x) }[/tex] with u; Then, we can have the above equation to be written as:
[tex]\mathbf{V = k_e \alpha \int^{-x+L}_{-x} \ \dfrac{(u+x) du}{\sqrt{u^2 +y^2 }} }[/tex]
By solving, we have:
[tex]\mathbf{V = k_e \alpha \int^{-x+L}_{-x} \ \dfrac{xdu}{\sqrt{u^2 + y^2 }}+ k_e \alpha \int ^{-x +2 }_{-x} \dfrac{udu}{\sqrt{u^2 + y^2 }} }[/tex]
[tex]\mathbf{V = \Big (( k_e \alpha x \ In(u+ \sqrt{u^2 +y^2})\Big )^{L-x}_{-x} +\Big(k_e\alpha \sqrt{u^2+y^2} \Big)^{L-x}_{-x} }[/tex]
[tex]\mathbf{V = k_e \alpha x \ In \Big (\dfrac{L-x + \sqrt{(L-x)^2+y^2 }}{-x+\sqrt{(-x)^2+y^2}} \Big)+ k_e \alpha \sqrt{(L-x)^2+y^2}- k_e \alpha \sqrt{(-x)^2+y^2}}[/tex]
Now, by replacing x with -d and y with 0, the coordinate of point B becomes [tex]\mathbf{x =\dfrac{L}{2} }[/tex] and [tex]\mathbf{y = b}[/tex].
Therefore, by solving the above equation, we can conclude that the electric potential at point B becomes;
[tex]\mathbf{V \Big (\dfrac{L}{2}, b \Big ) = k_e \alpha \dfrac{L}{2} \ In \Bigg ( \dfrac{\sqrt{(\dfrac{L}{2})^2+b^2}+ \dfrac{L}{2}} {\sqrt{(\dfrac{L}{2})^2+b^2}- \dfrac{L}{2}} \Bigg)}[/tex]
Learn more about the electrical potential at a point here:
https://brainly.com/question/21808222?referrer=searchResults
