Can you please help me answer all these questions for my homework 1) An object c
ID: 1410461 • Letter: C
Question
Can you please help me answer all these questions for my homework
1) An object carries a charge of -8.7 µC, while another carries a charge of -2.0 µC. How many electrons must be transferred from the first to the second object so that both objects have the same charge?
2) A plate carries a charge of -4 ?C, while a rod carries a charge of 2.50 ?C. How many electrons must be transferred from the plate to the rod, so that both objects have the same charge?
3) Two very small spheres are initially neutral and separated by a distance of 0.52 m. Suppose that 6.1 × 1013 electrons are removed from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere? (b) Is the force attractive or repulsive?
7) The masses of the earth and moon are 5.98 × 1024 and 7.35 × 1022 kg, respectively. Identical amounts of charge are placed on each body, such that the net force (gravitational plus electrical) on each is zero. What is the magnitude of the charge placed on each body?
8) A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 7400 N/C. The mass of the water drop is 2.67 × 10-9 kg. How many excess electrons or protons reside on the drop?
9) When point charges q1 = +7.2 ?C and q2 = +1.3 ?C are brought near each other, each experiences a repulsive force of magnitude 0.55 N. Determine the distance between the charges.
Explanation / Answer
1....The average excess charge = -8.7C - -2.0 µC / 2 = - 5.35 C .. this is the charge each will have when carrying equal charge.
Charge difference for -2.0C object = - - 5.35 C - (-2.0C) = - 3.35 C (by gaining N electrons) .. change differrence for -8.7C =- 5.35 C +8.7C = 3.35 C (by losing N electrons)
N x e = 3.35 C
N = 3.35 x 10^-6C / 1.60x 10^-19C .. .. 2.094 x 10^13 electrons transfer from -8.7C to -2C
2....The combined charge is
Q = (-4 C) + (2.5 C)
Q = -1.5 C
That charge equaly divided over the plate and the rod will give
q = (-1.5 C) / 2
q = -0.75 C
So you need to move
(-4 C) - (-0.75 C) = -3.25 C = -3.25 × 10^(-6) C
from the plate to the rod
1 electron has a charge of
Qe = -1.6 × 10^(-19) C
So the amount of electrons you need to move is
n = [ -3.25 × 10^(-6) C ] / [ -1.6 × 10^(-19) C ]
n = 20.3125 × 10^12 electrons
3.......a. Since each electron carries a charge of , the amount of negative charge removed from the first sphere is (6.1e13 electrons)( 1.6e-19 C / 1 electron)= 9.76e-6 C
Thus, the first sphere carries a charge +9.76 × 10-6 C, while the second sphere carries a charge -9.76 × 10-6 C. The magnitude of the electrostatic force that acts on each sphere is, therefore,
F= k (|q1|q2| / r^2) = (8.99e9)(9.76e-6)^2/(.52)^2 = 3.167N
b. Since the spheres carry charges of opposite sign, the force is:--> ATTRACTIVE.
7....Coulomb's law, force of attraction/repulsion
F = kQQ/r²
Q and Q are the charges in coulombs
r is separation in meters
k = 8.99e9 Nm²/C²
Gravitational attraction in newtons
F = G mm/r²
G = 6.67e-11 m³/kgs²
m and m are the masses of the two objects in kg
r is the distance in meters between their centers
set the two equal
kQQ/r² = G mm/r²
kQ² = G mm
Q = [ G mm/k ]
plug in the numbers. Note that your mass numbers are totally incorrect.
Please write other question seperately again. we are not autorised to answer multiple question in a single question but still i answered 4. Thanks hope you can understand my problem.