Electric field:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{k}\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}}$

**(a)**

Since both the charges are positive, the electric field will be zero at a point on the line joining them.

Let's take the distance from q_{1} (0.27 μC) to be x

Thus, the distance from q_{2} where the electric field will be zero is** d - x**, where d is the separation between the two charges.

The electric field due to charge q_{1} (0.27 μC) is:

$\overline{){{\mathbf{E}}}_{{\mathbf{1}}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}}{{\mathbf{x}}^{\mathbf{2}}}}$

Electric field due to charge q_{2} (0.71 μC) at d-x is:

$\overline{){{\mathbf{E}}}_{{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{(}\mathbf{d}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{2}}}}$

Point charges of 0.27 μC and 0.71 μC are placed 0.45 m apart.

a. At what point along the line between them is the electric field zero? Give your answer in m from 0.27 μC charge.

b. What is the magnitude of the electric field halfway between them in N/C?

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