Exercise – Electric Fields

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There are two identical rods of length 2a lying along the x-axis with their centers separated by a distance b (b > a). Each rod carries an equal charge of +Q uniformly distributed along their lengths. Please calculate the magnitude of the force exerted by the left rod on the right one.

Let the variable x_1 be the position placing on the left rod and the variable x_2 placing on the right rod. Here dx_1 (dx_2) is the infinitesimal displacement on the left (right) rod. The linear charge density is \lambda=Q/2a and the infinitesimal charge dq can be expressed as dq=\lambda dx_1. The force exerted on the left rod with an infinitesimal displacement dx_1 and an infinitesimal charge dq_1=\lambda dx_1 by an infinitesimal charge dq_2=\lambda dx_2 is:

    \[ dF=\frac{kdq_1dq_2}{(x_2-x_1)^2}=\frac{k\lambda^2dx_1dx_2}{(x_2-x_1)^2} \]

The total force is obtained through the integration:

    \[ F=k\lambda^2\int_{b-a}^{b+a}\left[\int_{-a}^a\frac{dx_1}{(x_1-x_2)^2}\right]dx_2 \]

In the inner integration, the x_1 is a variable and the x_2 is a constant so we have the equivalent relation dx_1=d(x_1-x_2). The above equation leads to:

    \begin{eqnarray*} F&=&k\lambda^2\int_{b-a}^{b+a}\left[\int_{-a}^a\frac{d(x_1-x_2)}{(x_1-x_2)^2}\right]dx_2 \\ &=&k\lambda^2\int_{b-a}^{b+a}\left[-\frac{1}{x_1-x_2}\right]_{x_1=-a}^adx_2 \\ &=&k\lambda^2\int_{b-a}^{b+a}\left(-\frac{1}{a-x_2}+\frac{1}{-a-x_2}\right)dx_2 \\ &=&k\lambda^2\int_{b-a}^{b+a}\left(\frac{1}{x_2-a}-\frac{1}{x_2+a}\right)dx_2 \\ &=&k\lambda^2\int_{b-a}^{b+a}\frac{1}{x_2-a}d(x_2-a)-k\lambda^2\int_{b-a}^{b+a}\frac{1}{x_2+a}d(x_2+a) \\ &=&k\lambda^2\left[\ln(x_2-a)\right]_{b-a}^{b+a}-k\lambda^2\left[\ln(x_2+a)\right]_{b-a}^{b+a} \\ &=&k\lambda^2\ln(\frac{b}{b-2a})-k\lambda^2\ln(\frac{b+2a}{b}) \\ &=&k\lambda^2\ln(\frac{b^2}{(b-2a)(b+2a)}) \end{eqnarray*}

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