A +2.00nC point charge is at the origin, and a second -5.00nC point charge is on

Question

A +2.00nC point charge is at the origin, and a second -5.00nC point charge is on the x-axis at x= 0.800m.

Part A

Find the magnitude of the electric field at x = 0.200m on the x-axis.

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Part B

Find the direction of the electric field at x = 0.200m on the x-axis.

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A +2.00nC point charge is at the origin, and a second -5.00nC point charge is on the x-axis at x= 0.800m.

Part A

Find the magnitude of the electric field at x = 0.200m on the x-axis.

E =   N/C

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Part B

Find the direction of the electric field at x = 0.200m on the x-axis.

= counterclockwise from the +x-direction

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Explanation / Answer

As the two charges have opposite signs their fields directions are the same at X .. (out from +ve, inwards to -ve charge), so their fields ADD at point X (this accounts for their opposite charge)

Field strength = kq/d²
Net field .. E = {k*2.0^-9C / (0.20m)²} + {k*5.0^-9C / (0.8m-0.2m)²} .. (k = 9.0^9)

E = 18/0.04 + 45/0.36 .. .. E = 575.0 N/C

Field direction .. as described in my first sentence .. from + to - charge (in +x direction)

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