Question: Figure 24-47 shows a thin plastic rod of length L = 12.0 cmand uniform positive charge Q = 56.1fClying on an xaxis. With V = 0at infinity, find the electric potential at point P₁ on the axis, at distance d = 250 cmfrom the rod.

1. Length of the rod, L = 12cm
2. Uniform positive charge on the axis, Q = 56.1fC
3. The electric potential at V = 0 is infinity.
4. The distance of the point, d = 2.50 cm

Substituting the value of charge from the concept of the linear charge density in the formula of the electric potential due to a charge at a point, we can get the required value of the potential on the x-axis by using the given data.

Formulae:

The linear charge density of a distribution, $\lambda =\frac{dq}{dx}$ (i)

The electric potential at a point due to a point charge, $dV=\frac{1}{4\pi {\epsilon }_0}\left(\frac{dq}{x+d}\right)$ (ii)

Consider an infinitesimal segment dx that has a charge.

The expression for the electric potential due to the small element of the finite line of charge along the direction of the line charge is given using the equation (i) and then the given values in equation (ii) as follows:

$dV=\frac{1}{4\pi {\epsilon }_0}\left(\frac{\lambda dx}{x+d}\right)$

Now, the electric potential is given by integrating the above value and substituting the given values as follows:(Here, the length of the charge is changing from 0 to L)

Therefore, the value of the potential at point P is 7.396 x 10^-3V