plz do all of them (7%) Problem 2: Four point charges of equal magnitude Q = 35
ID: 1656136 • Letter: P
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plz do all of them
(7%) Problem 2: Four point charges of equal magnitude Q = 35 nC are placed on the corners of a rectangle of sides D , = 23 cm and D =5 cm. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative. Refer to the figure. Q Otheexpertta.com a aWhich hand 14% Part () of the following represents a free-body diagram for the charge on the lower left corner of the rectangle? a 14% Part (b) Enter an expression for the horizontal component of the net force acting on the charge located at the lower left corner of the rectangle in terms of the charges, the given distances, and the Coulomb constant. Use a coordinate system in which the positive direction to the right. Grade Summary Deductions Potential 100 Submissions Attempts remaining:3 (% per attempt detailed view k) (D +D) submit Him Tete poetwo Hints: 3% deduction per hint. Hints remaining 3 Feedback : 3% deduction per feedback. D A 14% Part (c) Calculate the value of the horizontal component of the net force, in newtons, on the charge located at the lower left corner of the rectangle. A 14% Part (d) Enter and expression for the vertical component of the net force acting on the charge located at the lower left corner of the rectangle in terms of the charges, the given distances, and the Coulomb constant. Assume up is positive. A 14% Part (e) Calculate the value of the vertical component of the net force, in newtons. A 14% Part (O ) Calculate the magnitude of the net force, in newtons, on the charge located at the lower left corner of the rectangle. A 14% Part (8 Calculate the angle, in degrees between -180° and +180°, that the net force makes, measured counterclockwise from the positive horizontal direction.Explanation / Answer
given,
Q = 35 nC
D1 = 23 cm
D2 = 5 cm
electric force = k * q1 * q2 / distance^2
horizontal distance of upper left and lower left charge = 0 cm
so,
horizontal force between upper left and lower left charge will be 0
horizontal force between lower left and lower right charge will be = k * Q * Q / D1^2
horizontal force between lower left and upper right charge will be = k * Q * Q * cos(theta) / (sqrt(D1^2 + D2^2))^2
cos(theta) = D1 / sqrt(D1^2 + D2^2)
horizontal force between lower left and upper right charge will be = k * Q * Q * D1 / (D1^2 + D2^2)^(3/2)
total horizontal force on lower left charge = k * Q * Q / D1^2 + k * Q * Q * D1 / (D1^2 + D2^2)^(3/2)
b) total horizontal force on lower left charge = k * Q^2(1 / D1^2 + D1 / (D1^2 + D2^2)^(3/2))
putting values in equation we'll get
total horizontal force = 9 * 10^9 * (35 * 10^-9)^2 * (1 / (0.23)^2 + 0.23 / (0.23^2 + 0.05^2)^(3/2))
c) total horizontal force = 0.0004028 N or 4.028 * 10^-4 N
vertical force between lower left and upper left charge will be = k * Q * Q / D2^2
vertical force between lower left and lower right charge will be = 0 N
vertical force between lower left and upper right charge will be = k * Q * Q * sin(theta) / (sqrt(D1^2 + D2^2))^2
sin(theta) = D2 / sqrt(D1^2 + D2^2)
vertical force between lower left and upper right charge will be = k * Q * Q * D2 / (D1^2 + D2^2)^(3/2)
total vertical force on lower left charge = k * Q * Q / D2^2 - k * Q * Q * D2 / (D1^2 + D2^2)^(3/2)
d) total vertical force on lower left charge = k * Q^2(1 / D2^2 + D2 / (D1^2 + D2^2)^(3/2))
putting values in equation we'll get
total vertical force = 9 * 10^9 * (35 * 10^-9)^2 * (1 / (0.05)^2 + 0.05 / (0.23^2 + 0.05^2)^(3/2))
e) total vertical force = 0.004452 N or 4.452 * 10^-3 N
net force on lower left charge = sqrt(hozizontal force^2 + vertical force^2)
net force on lower left charge = sqrt(0.0004028^2 + 0.004452^2)
f) net force on lower left charge = 0.00447018 N or 4.47018 * 10^-3 N
direction = tan^-1(vertical force / horizontal force)
direction = tan^-1(0.004452 / 0.0004028)
g) direction = 84.83 degree
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