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.
- billrussell42Lv 71 month agoFavorite 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.
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
E = F/Q = kQ/r²
Q = F/E F = QE
in Newtons/coulomb OR volts/meter
k = 1/4πε₀ = 8.99e9 Nm²/C²