Anonymous asked in Science & MathematicsPhysics · 1 month ago

e-field magnitude?

Four charges reside on the corners of a square. Three of the charges are positive: 4μC, 1μC, and 4μC. The 4μC charges are at diagonal corners. The final charge is negative: -3μC. Determine the magnitude of the E-field found at the center of the square if each side has a length of 2.0 m.

20091 N/C

63002 N/C

55210 N/C

18108 N/C

2299 N/C

1 Answer

  • 1 month ago
    Favorite Answer

    add up the force due to each as vectors.

    diagonal of the square is 2√2, and distance of each from the center is √2

    put the center of the square at the origin, the 4µC ones on the +x and –x axis at ±√2

    the +1 on the +y axis, the –3 on the –y axis, both at √2.

    E due to the 4µC charges, the fields cancel

    E due to the +1µC

    E₁x = 0

    E₁y = k(1µ)/(√2)² = (9e9)(1e-6)/2 = (9/2)e3 = 4500 N/C pointed down

    E due to the –3µC

    E₂x = 0

    E₂y = k(3µ)/2 = (27/2)e3 = 13500 N/C pointed down

    total of the two is 18000 N/C pointed down ⬅

    18108 is closest, although the numbers don't justify that precision.

     If I used a more accurate value for k, the answer would be 17980 N/C. So I don't know where the 18108 came from.

    Electric field

    The strength or magnitude of the field at a given point

    is defined as the force that would be exerted on a

    positive test charge of 1 coulomb placed at that point;

    the direction of the field is given by the direction of

    that force.

       E = F/Q = kQ/r²

       Q = F/E F = QE

    in Newtons/coulomb OR volts/meter

       k = 1/4πε₀ = 8.99e9 Nm²/C²

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