# What is the strength of the electric field at the position indicated by the dot in the figure? phot

What is the strength of the electric field at the position indicated by the dot in the figure?
photo
What is the strength of the electric field at the position indicated by the dot in the figure?
$E=\overline{\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}}\frac{N}{C}$
What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.
$\theta ={\overline{\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}}}^{\circ }$ .
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Step 1
Since the charges are at equal distances and both are positive, they will repel with the same magnitude. If you draw the forces, then they will cancel out the y component of the final field. There will only be the x component. And since they are the same, then the final electric field will be twice the x component of one electric field of a charge.
$E=\frac{kq}{{r}^{2}}$

$E=\frac{\left(9×{10}^{9}\right)\left(1×{10}^{-9}\right)}{{0.0707}^{2}}$
$=1800.5\frac{N}{C}$
Now to find the x component,
$\mathrm{cos}45=\frac{x}{1800.5}$
$x=1273.18\frac{N}{C}$
Now just multiply by 2 to accomodate both charges,
$E=2546.35\frac{N}{C}$ in direction of 0 degrees (no y component since they cancel each other out)