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July 30, 2014

July 30, 2014

Posted by **lisa** on Saturday, May 28, 2011 at 5:48pm.

Two equal negative charge on 9 nC each are in the water (ε = 81) at a distance of 8 inches apart. Determine the strength and potential of the field at a point located at a distance of 5 cm from the charges

- physics -
**Damon**, Saturday, May 28, 2011 at 6:16pm8 cm apart I assume

Do the potential first. It is easy because you can add potentials due to each charge without doing any geometry.

V = 2 * [ q/ (4 pi eo r ) ]

for each E field you must first find the E vector away from each charge and multiply by the cosine of the angle between that and the direction perpendicular to the line between the charges

Draw the triangle

base 8, legs 5

half the base = 4

right triangles, 3, 4 ,5

cos A = 3/5

so

E = 2 [ q/(4 pi eo r^2)](3/5)

- physics -
**jackie**, Sunday, May 29, 2011 at 10:23amexplanation on how to use Snell's law for refraction to calculate the refrection

- physics -
**Anonymous**, Thursday, November 17, 2011 at 4:15amA ball of mass 8kg fall from rest from a height of 100m calculate the k.E energy after falling with a distance of 30m

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