Question

A straight line segment of length L (see Figure below) carries a uniform line charge
λ. Find the electric field a distance z above one end of the straight line. Indicate the
direction of the electric field vector. Check that your formula is consistent with what
you would expect for the case z  L.

Answer 1

Answer on Question #81290, Physics / Electromagnetism

Question:

A straight line segment of length $L$ (see Figure below) carries a uniform line charge $\lambda$. Find the electric field a distance $z$ above one end of the straight line. Indicate the direction of the electric field vector. Check that your formula is consistent with what you would expect for the case $zL$.

Solution:

In the point $P$ a part of $L$ with the length $dx$ gives $dE = \frac{kq}{(z + L - x)^2} = \frac{k\lambda dx}{(z + L - x)^2}$, therefore according to superposition principle the electric field vector modulus

$$E = \int_{0}^{L} \frac{k\lambda dx}{(L + z - x)^2} = \frac{kL\lambda}{(z + L)z}.$$

In case of $z \square L$, $E = \frac{kq}{z^2}$, where $q = \lambda L$.

If $\lambda > 0$ the direction of $E$ is from the segment, otherwise – toward it.

The answer:

$$E = \int_{0}^{L} \frac{k\lambda dx}{(L + z - x)^2} = \frac{kL\lambda}{(z + L)z}.$$

In case of $z \square L$, $E = \frac{kq}{z^2}$, where $q = \lambda L$.

If $\lambda > 0$ the direction of $E$ is from the segment, otherwise – toward it.

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