## Related questions with answers

If the total charge on a thin rod of length 0.4 m is 2.5 nC, what is the magnitude of the electric field at a location 1 cm from the midpoint of the rod, perpendicular to the rod?

Solution

VerifiedIn this problem we evaluate the electric field of a thin rod of length $L=40$ cm and total charge $Q=2.5$ nC at a point $x=1$ cm away from the midpoint. We know that the field points perpendicular to the rod, which is the $x$ direction here and we use the formula for the field at the point $x$ away from the midpoint

$\begin{align*} E_x&=\frac{1}{4\pi\varepsilon_0}\qty(\frac{Q}{x\sqrt{x^2+(L/2)^2}})\\ &=\frac{1}{4\pi(8.85\cdot10^{-12}\text{ N}^{-1}\text{C}^2\text{m}^2)}\qty(\frac{2.5\cdot10^{-9}\text{ C}}{(0.01\text{ m})\sqrt{(0.01\text{ m})^2+(0.4\text{ m}/2)^2}})\\ &=11220.42\text{ N/C}\\ &=\boxed{\color{#c34632}1.12\cdot10^4\text{ N/C}.} \end{align*}$

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