Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Physics Electric Field Help!?
Two point charges of -2.00(micro Coulombs) and +5.00 (micro Coulombs) are placed along the x-axis 0.200 m apart. What is the electric field at point P which is 0.200m vertically away from the mid-point separating the charges
- oldschoolLv 72 months agoFavorite Answer
Assume the -2µC charge is at the origin. The +5µC charge is at (0.2,0).
Point P is at (0.1,0.2). The distance r² from each charge to
P = 0.1²+0.2² = 0.05 = 1/20
The E field at P from the -2µC = k*2e-6/0.05 = k*40e-6 at 243.435° = E2
The E field at P from the +5µC = k*2e-6/0.05 = k*10e-5 at 116.565° = E5
E2 = 359,502 N/C
E5 = 898,755 N/C
Sum the horizontal components:
E2*cos243.435 + E5*cos116.565 = -562,709 N/C
Sum the vertical components:
E2*sin243.435 + E5*sin116.565 = 482,323 N/C
741,132 N/C at 139.4°
or about 741 kN/C at 139° <<<<<
- az_lenderLv 72 months ago
If the -2 uC charge is to the left of the +5 uC charge, the field due to the lefthand charge is
-161.0 kN/C i - 322.0 kN/C j
and the field due to the righthand charge is
-402.5 kN/C i + 805.0 kN/C j.
Then the total field at P is
-563.5 kN/C i + 483.0 kN/C j,
which has a magnitude of 742.2 kN/C.
I needn't go any further about the direction, because they never actually told us whether P was above or below the x-axis, and they never actually said whether the -2 uC charge was to the left or to the right of the +5 uC charge.