# Find (a) the magnitude and (b) the direction of the net electric field at the origin.?

Two charges are located on the x axis: q1 = +5.6C at x1 = +3.4 cm, and q2 = +5.6C at x2 = -3.4 cm. Two other charges are located on the y axis: q3 = +4.0C at y3 = +5.3 cm, and q4 = -8.5C at y4 = +7.7 cm. Find (a) the magnitude and (b) the direction of the net electric field at the origin.

### 1 Answer

- NCSLv 73 months agoFavorite Answer
field E = vector Σ kQ / d²

where k = 8.99e9 N·m²/C²

x-direction:

The two charges on the x-axis have the same magnitude and same distance from the origin; therefore the field at the origin due to these two charges is 0 N/C.

y-direction:

Ey = 8.99e9N·m²/C² * [-4.0C/(0.053m)² - 8.5C/(0.077m)²]

Both terms in the [] have negative signs since their effect on a positive charge at the origin would be to pull/push it down.

Ey = -2.6e13 N/C

and since Ex = 0 as explained above,

E = 2.6e13 N/C down.

Now, 5.6 C is an AWFUL lot of charge. Typically charges in this type of problem are expressed in µC (1e-6), nC (1e-9) or pC (1e-12). If your charges were actually in µC, then multiply my answer by 1e-6.

If you find this helpful, please award Best Answer. You get points too!

Ex = 8.99e9N·m²/C² * (-5.6C - 5.6C) / (

You're welcome. Thanks for awarding!