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(A) The electric field $E$ due to a point charge $q$ at a distance $r$ is given by $E = k|q|/r^2$. For two identical positive charges at $x_1 = 2$ and

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(A) The electric field $E$ due to a point charge $q$ at a distance $r$ is given by $E = k|q|/r^2$. For two identical positive charges at $x_1 = 2$ and $x_2 = 4$, the net electric field $E_{net}$ at any point $x$ is the vector sum of the fields from both charges. $1$. Near the charges ($x 	o 2$ or $x 	o 4$), the field magnitude approaches infinity because $r 	o 0$. $2$. At the midpoint $x = 3$, the electric field from the charge at $x = 2$ is $E_1 = kq/(3-2)^2 = kq$ (pointing in the $+x$ direction), and the field from the charge at $x = 4$ is $E_2 = kq/(4-3)^2 = kq$ (pointing in the $-x$ direction). $3$. Thus, at $x = 3$, $E_{net} = E_1 - E_2 = 0$. $4$. The graph must show a singularity (asymptote) at $x = 2$ and $x = 4$, and a zero at $x = 3$. Comparing this with the given options, the graph that shows $E 	o \infty$ at $x=2$ and $x=4$ and $E=0$ at $x=3$ is the correct one.

Two charged particles, each with a charge of $+q$, are located along the $x$-axis at $x = 2$ and $x = 4$, as shown below. Which of the following shows the graph of the magnitude of the electric field along the $x$-axis from the origin to $x = 6$?

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