Question

On a thin rod of length L lying along the x-axis with one end at the origin (x = 0), there is distributed a charge per unit length given by λ = b x, where b
is a constant.
(a) Taking the electrostatic potential at infinity to be zero, find V at the point P on the
y-axis.
(b) Determine the vertical component, Ey, of the electric field intensity at P from the
result of part (a).
(c) What is the horizontal component, Ex, of the electric field intensity at point P?

Answer 1

Answer 2

Calculations:

(a) Finding $V$ at Point $P$:

1. Tiny Charge Piece:
   Imagine chopping the rod into tiny bits of length $dx$, each at position $x$. The charge on one bit is:

   $$dq=\lambda dx=bxdx$$

2. Potential from One Bit:
   The potential at $P$ (distance $r=\sqrt{x^2+y^2}$) due to $dq$ is:

   $$dV=\frac{1}{4\pi {ϵ}_0}\frac{dq}{r}=\frac{bxdx}{4\pi {ϵ}_0\sqrt{x^2+y^2}}$$

3. Add Up All Bits (Integrate):
   Sum contributions from $x=0$ to $x=L$:

   $$V=\frac{b}{4\pi {ϵ}_0}{\int }_0^L\frac{xdx}{\sqrt{x^2+y^2}}$$

    

   Final result:

   $$V=\frac{b}{4\pi {ϵ}_0}(\sqrt{L^2+y^2}-\mathrm{\mid }y\mathrm{\mid })$$

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(b) Finding $E_y$ from $V$:

1. Electric Field as Slope of $V$:
   $E_y=-\frac{dV}{dy}$. Differentiate the result from (a):

   $$\frac{d}{dy}(\sqrt{L^2+y^2}-\mathrm{\mid }y\mathrm{\mid })=\frac{y}{\sqrt{L^2+y^2}}-\text{sgn}(y)$$

   (Here, $\text{sgn}(y)=+1$ if $y>0$, $-1$ if $y<0$).

2. 
   

3. Final Expression:

   $$E_y=-\frac{b}{4\pi {ϵ}_0}(\frac{y}{\sqrt{L^2+y^2}}-\text{sgn}(y))=\frac{b}{4\pi {ϵ}_0}(\text{sgn}(y)-\frac{y}{\sqrt{L^2+y^2}})$$

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(c) Why $E_x=0$:

• Symmetry Says:
  For every charge bit at $+x$, there’s an equal bit at $-x$ (if the rod extended symmetrically). Their $E_x$ components cancel.

• Math Says:
  $V$ doesn’t depend on $x$, so $E_x=-\frac{\mathrm{\partial }V}{\mathrm{\partial }x}=0$.

Answers:

• (a) Potential $V$ at $P$:

  $$V=\frac{b}{4\pi {ϵ}_0}(\sqrt{L^2+y^2}-\mathrm{\mid }y\mathrm{\mid })$$

• (b) Vertical field $E_y$:

  $$E_y=\frac{b}{4\pi {ϵ}_0}(\text{sgn}(y)-\frac{y}{\sqrt{L^2+y^2}})$$

• (c) Horizontal field $E_x$:

  $$E_x=0$$