1) A 8.01 nC charge is located 1.69 m from a 4.54 nC point charge. (a) Find the

Question

1) A 8.01 nC charge is located 1.69 m from a 4.54 nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other

2) A small sphere of mass m = 6.30 g and charge q1 = 27.1 nC is attached to the end of a string and hangs vertically as in the figure. A second charge of equal mass and charge q2 = 58.0 nC is located below the first charge a distance d = 2.00 cm below the first charge as in the figure

(a) Find the tension in the string.

3) A molecule of DNA (deoxyribonucleic acid) is 2.10 µm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.25% upon becoming charged, and remains in this equilibrium position. Determine the effective spring constant of the molecule.

4) A positive charge q1 = 2.60 µC on a frictionless horizontal surface is attached to a spring of force constant k as in the figure shown below. When a charge of

q2 = 8.750 µC is placed 9.50 cm away from the positive charge, the spring stretches by 5.00 mm, reducing the distance between charges to d = 9.00 cm. Find the value of k.

(b) If the string can withstand a maximum tension of 0.180 N, what is the smallest value d can have before the string breaks?
cm

5) Three point charges are located at the corners of an equilateral triangle as in the figure below. Find the magnitude and direction of the net electric force on the 0.80 µC charge. (Let A = 0.80 µC, B = 7.30 µC, and C = 3.80 µC.) what is the magnitude and direction?

Explanation / Answer

1)

F=Q1*Q2 / (4 * pi * Eo * r^2)

Q1= 8.01 * 10^-9

Q2= 4,54 * 10^-9

pi=3.14

r= 1.69 m

Eo = 8.854 x 10^-12

therefore F =( 8.01 * 4.54 * 10^-18 ) / ( 4* 3.14 * 8.85 ×10^12 * 1.69^2 )

F = 11.45* 10^ -8 N

2)

the weight of the sphere acting downward = mg = 6.30*10^-3 * 9.81 N = 0.06180 N
The attractive force on the charge due to negative charge = 9*10^9*27.1*10^-9*58*10^-9/(0.02)^2 N
Simplifying using calculator, F down = 3793.3 * 10^-5 = 37.933*10^-3 N
Hence the total force downward = 99.733*10^-3 N

3)

The ends of the molecules become singly ionized meaning the charge on both ends is e.

e = 1.6x10^-19 C

I will make an assumption that the force stays constant during compression.

F = kQ²/r²
F = (9x10^9)(1.6x10^-19)²/(2.10x10^-6)²
F = 5.22x10^-17 N

F = Kx
x = 0.0125(2.10x10^-6)

5.22x10^-17 = K(0.0125(2.10x10^-6)
K = 1.98*10^-9 N/m

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