Question

Two test charges are located in the x–y plane. If q1=−3.05 nC and is located at x1=0.00 m, y1=0.640 m, and the second test charge has magnitude of q2=3.80 nC and is located at x2=1.50 m, y2=0.800 m, calculate the x and y components, Exand Ey, of the electric field →E in component form at the origin, (0,0). The Coulomb force constant is 1/(4πϵ0)=8.99×109 N⋅m2/ C2.

Answer #1

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Electric field direction is radially outwards for a positive charge and it is radially inwards for a negative charge.

Electric field E_{2} is acting at an angle, split it
into its components

Consider point P

Consider all horizontal electric fields

**ANSWER:
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Consider all vertical electric fields

**ANSWER:
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Magnitude

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Direction

in 2nd quadrant

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