Questions: Problem 7: (11% of Assignment Value) A thin plastic rod of length d=2.4 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform charge density of λ=8.5 nC / m. The point P is located along the y-axis at (0, y=25 cm). Part (a) Enter an expression for the electric potential at point P due to a small segment of the rod, of thickness d x, located at (x, 0). Give a symbolic answer in terms of λ, x, d x, y, and the Coulomb constant ke. d V= Part (b) Using the expression you entered in part (a), choose the correct limits for the integral that will give the electric potential at P. Part (c) Calculate the electric potential at P, in volts.

Solution Steps

To find the electric potential \( dV \) at point \( P \) due to a small segment of the rod, we start by considering a small segment of thickness \( dx \) located at \( (x, 0) \) on the rod. The charge of this small segment is \( dq = \lambda \, dx \).

The distance \( r \) from this segment to point \( P \) located at \( (0, y) \) is given by: \[ r = \sqrt{x^2 + y^2} \]

The electric potential \( dV \) due to this small segment is: \[ dV = \frac{k_e \, dq}{r} = \frac{k_e \, \lambda \, dx}{\sqrt{x^2 + y^2}} \]

Thus, the expression for the electric potential at point \( P \) due to a small segment of the rod is: \[ dV = \frac{k_e \, \lambda \, dx}{\sqrt{x^2 + y^2}} \]

\[ \boxed{dV = \frac{k_e \lambda \, dx}{\sqrt{x^2 + y^2}}} \]

The rod lies along the \( x \)-axis with its midpoint at the origin, and its total length is \( d = 2.4 \, \text{m} \). Therefore, the rod extends from \( -1.2 \, \text{m} \) to \( 1.2 \, \text{m} \).

The correct limits for the integral to find the total electric potential at point \( P \) are from \( -1.2 \, \text{m} \) to \( 1.2 \, \text{m} \).

\[ \boxed{-1.2 \, \text{m} \text{ to } 1.2 \, \text{m}} \]

To find the total electric potential \( V \) at point \( P \), we integrate the expression for \( dV \) over the length of the rod:

\[ V = \int_{-1.2}^{1.2} \frac{k_e \lambda \, dx}{\sqrt{x^2 + y^2}} \]

Given:

• \( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)
• \( \lambda = 8.5 \times 10^{-9} \, \text{C} / \text{m} \)
• \( y = 0.25 \, \text{m} \)

\[ V = k_e \lambda \int_{-1.2}^{1.2} \frac{dx}{\sqrt{x^2 + 0.25^2}} \]

This integral can be evaluated using a standard integral formula:

\[ \int \frac{dx}{\sqrt{x^2 + a^2}} = \ln \left( x + \sqrt{x^2 + a^2} \right) + C \]

Applying the limits:

\[ V = k_e \lambda \left[ \ln \left( x + \sqrt{x^2 + 0.25^2} \right) \right]_{-1.2}^{1.2} \]

\[ V = k_e \lambda \left[ \ln \left( 1.2 + \sqrt{1.2^2 + 0.25^2} \right) - \ln \left( -1.2 + \sqrt{(-1.2)^2 + 0.25^2} \right) \right] \]

Since \( \ln(a) - \ln(b) = \ln \left( \frac{a}{b} \right) \):

\[ V = k_e \lambda \ln \left( \frac{1.2 + \sqrt{1.2^2 + 0.25^2}}{-1.2 + \sqrt{1.2^2 + 0.25^2}} \right) \]

Calculating the values inside the logarithm:

\[ \sqrt{1.2^2 + 0.25^2} = \sqrt{1.44 + 0.0625} = \sqrt{1.5025} \approx 1.2258 \]

\[ V = 8.99 \times 10^9 \times 8.5 \times 10^{-9} \times \ln \left( \frac{1.2 + 1.2258}{-1.2 + 1.2258} \right) \]

\[ V = 76.415 \times \ln \left( \frac{2.4258}{0.0258} \right) \]

\[ V = 76.415 \times \ln (94.0078) \]

\[ V = 76.415 \times 4.5439 \approx 347.4 \, \text{V} \]

\[ \boxed{V \approx 347.4 \, \text{V}} \]

Final Answer