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} 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} 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} at r = 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 25.0 µC and 45.0 µC are placed 0.500 m apart.

(a) At what point along the line between them is the electric field zero?

(b) What is the electric field halfway between them?

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