Question

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?

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

Ex=14πε0(Qxx2+(L/2)2)=14π(8.851012 N1C2m2)(2.5109 C (0.01 m )(0.01 m )2+(0.4 m /2)2)=11220.42 N/C=1.12104 N/C.\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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