# A rod 7.5 cm long is uniformly charged and has a total charge of - 13.1 \mu C. Find the magnitude...

## Question:

A rod {eq}7.5 cm {/eq} long is uniformly charged and has a total charge of {eq}- 13.1 \mu C {/eq}. Find the magnitude of the electric field along the axis of the rod at a point {eq}18.271 cm {/eq} from the center of the rod.

The Coulomb constant is {eq}8.98755 \times 10^9 N \cdot m^2/C^2. {/eq}

## Electric Field from Charge Densities:

The electric field due to a point charge can be found using Coulomb's Law: {eq}E = \dfrac {kq}{r^2} {/eq}, where *k* is the Coulomb's constant, *q* is the charge magnitude, and *r* is the distance from the charge to the point of observation. In the case of the system of charges, the total field is the vector sum of all individual fields from each charge in the system.

## Answer and Explanation:

**Given:**

- {eq}L = 0.075 \ m {/eq} is the length of the rod;

- {eq}d = 0.18271 \ m {/eq} is the distance from the center of the rod;

- {eq}k = 8.98755 \times 10^9 \ N \cdot m^2/C^2 {/eq} is the Coulomb's constant;

- {eq}q = -1.31\times 10^{-5} \ C {/eq} is the charge on the rod;

The electric field at the point distance *d* from the center of the rod is:

{eq}E = \displaystyle \int \limits_0^L \dfrac {kq}{L} \cdot \dfrac {dx}{\left ( \dfrac {L}{2} + d - x \right )^2} {/eq}

Computing the integral and simplifying, we obtain:

{eq}E = \dfrac {kq}{L} \left ( \dfrac {1}{d - \dfrac {L}{2}} - \dfrac {1}{d + \dfrac {L}{2}} \right ) = \dfrac {4kq}{4d^2 - L^2} {/eq}

Calculation yields:

{eq}E = \dfrac {4\cdot 8.98755 \times 10^9 \ N \cdot m^2/C^2 \cdot (-1.31\times 10^{-5} \ C)}{4 (0.18271 \ m)^2 - (0.075 \ m)^2} \approx 3.68\times 10^6 \ N/C {/eq}

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