Electric Fieldss
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A 43.2 nC Charge is located at position (x,y) = (1.0 cm, 2.0 cm). At what position (x,y) position is the electric field
1) At -225,000i N/C?
2) At (161,000i + 80,500j) N/C
X=?
Y=?
3) At (21,000i - 28,800j) N/C
X=?
Y=?
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Solution Summary
A 43.2 nC Charge is located at position (x,y) = (1.0 cm, 2.0 cm). At what position (x,y) position is the electric field
Solution Preview
The absolute value of the electric field is
E = kq/Δr^2, (1)
where k = 8.988×10^9 N m^2 C^{−2}, the charge is given as q = 43.2×10^{-9} C, and Δr is the distance from the charge (NOT from the origin).
The absolute values of the electric field in the 3 cases given are
E_1 = 225,000 N/C,
E_2 = sqrt(161,000^2 + 80,500^2) ≈ 180,000 N/C,
E_3 = sqrt(21,000^2 + 28,800^2) ≈ 35,643 N/C.
From equation (1) we have
Δr ...
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