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adca14

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## Homework Statement

Three charges, +2.5[tex]\mu[/tex]C, -4.8[tex]\mu[/tex]C & -6.3[tex]\mu[/tex]C

are located at (-0.20m, 0.15m), (0.50m, -0.35m) and (-0.42m, -0.32m) respectively. What is the electric field at the origin?

q1 = +2.5[tex]\mu[/tex]C

q2 = -4.8[tex]\mu[/tex]C

q3 = -6.3[tex]\mu[/tex]C

## Homework Equations

a[tex]^{}2[/tex] + b[tex]^{}2[/tex] = c[tex]^{}2[/tex]

v[tex]_{}x[/tex] = magnitude [tex]\times[/tex]cos([tex]\theta[/tex])

v[tex]_{}y[/tex] = magnitude [tex]\times[/tex]sin([tex]\theta[/tex])

E = [tex]\frac{kq}{r^{}2}[/tex]

law of cosines

k= 9x10^9

## The Attempt at a Solution

first i found the hypotenuse for the three charges.

q1 = .25m

q2 = .6103m

q3 = .5280m

then i used the formula for magnitude of an electric field

where k is the constant, q was my three charges, and the radius were my three hypotenuses

my results were:

q1 = 360000

q2 = 115983

q3 = 203383

i used the law of cosines to get [tex]\theta[/tex]

my three angles were:

1 = 36.8

2 = 35

3 = 37.3

to find Ex, i multiplied the product of my magnitudes by the cosine of its respective angle:

my results:

1 = 288263

2 = 95007

3 = 161785

i added these up and got 545055, the book says 2.2x10^5!

i didnt bother doing y, since I am completely lost!

Please help me!