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Anonymous asked in Science & MathematicsPhysics · 2 months ago

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 

2 Answers

  • 2 months ago
    Favorite 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° <<<<<

  • 2 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.

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