A charge Q = +4.5 uC is located at the origin and is fixed so that it cannot move. A small object with mass m = 2.0 g and charge q = +8.0 uC is released from rest at point A on the x-axis at x = 2.0 cm. The small object moves without friction along the x-axis under the influence of the charge Q only (gravity can be neglected) and eventually reaches point B on the x-axis at x = 10.0 cm. (a) Calculate the potential difference between point B and point A (change in VAB = VB-VA). (b) Calculate the speed of the small object when it reaches point B
Δφ=φ₁-φ₂=kQ/r₁-kQ/r₂=
=kQ(r₂-r₁)/r₁•r₂=
=9•10⁹•4.5•10⁻⁶•8•10⁻²/20•10⁻⁴=1.62•10⁶ V
mv²/2= q•Δφ,
v= sqrt(2•q•Δφ/m)=…
Elena,
Im a little stuck on the first equation you have given. In the denominator, we are supposed to multiply r1 by r2. How did you get 20 x 10? Shouldn't it be 0.1m x 0.02m?
Shouldn't the answer be 1.62*10^4V? Unless i'm putting the numbers in wrong. Please help
To find the potential difference between points A and B (VB-VA), we can use the formula:
ΔV = k(Q/rB - Q/rA)
Where:
- ΔV is the potential difference
- k is the Coulomb's constant (k = 8.99 x 10^9 N m^2/C^2)
- Q is the charge at the origin (+4.5 uC = 4.5 x 10^-6 C)
- rB is the distance between B and the origin (rB = 10 cm = 0.1 m)
- rA is the distance between A and the origin (rA = 2 cm = 0.02 m)
Substituting the given values into the formula:
ΔV = (8.99 x 10^9 N m^2/C^2) * [ (4.5 x 10^-6 C) / (0.1 m) - (4.5 x 10^-6 C) / (0.02 m) ]
Simplifying the expression:
ΔV = (8.99 x 10^9 N m^2/C^2) * (4.5 x 10^-6 C) * (1/0.1 - 1/0.02)
ΔV = (8.99 x 10^9 N m^2/C^2) * (4.5 x 10^-6 C) * (10 - 50)
ΔV = (8.99 x 10^9 N m^2/C^2) * (4.5 x 10^-6 C) * (-40)
ΔV = -0.16191 J
Therefore, the potential difference between point B and point A is approximately -0.16191 J. The negative sign indicates that the potential decreases as the object moves from A to B.
To calculate the speed of the object when it reaches point B, we can consider the conservation of energy.
The total mechanical energy at point A is given by:
Ei = KE + PE
Since the object starts from rest at point A, the initial kinetic energy (KEi) is 0.
Ei = PE
The potential energy (PE) at point A is given by:
PE = q * ΔV
Where:
- q is the charge of the object (+8.0 uC = 8.0 x 10^-6 C)
- ΔV is the potential difference from part (a)
Substituting the values:
PE = (8.0 x 10^-6 C) * (-0.16191 J)
PE = -1.29528 x 10^-6 J
The total mechanical energy at point B (Ef) is given by:
Ef = KE + PE
Since the object has no potential energy at point B, only kinetic energy remains:
Ef = KE
Let's assume the final speed at point B is vB.
Ef = KEf = 0.5 * m * vB^2
Since the object has no initial kinetic energy and no potential energy at point B, the total mechanical energy is conserved:
Ei = Ef
PE = 0.5 * m * vB^2 - 1.29528 x 10^-6 J
Now we can solve for vB:
0.5 * m * vB^2 = 1.29528 x 10^-6 J
Substituting the given values:
0.5 * (2.0 g) * (vB^2) = 1.29528 x 10^-6 J
Converting the mass from grams to kilograms:
0.5 * (0.002 kg) * (vB^2) = 1.29528 x 10^-6 J
Simplifying the expression:
0.001 * (vB^2) = 1.29528 x 10^-6 J
vB^2 = (1.29528 x 10^-6 J) / (0.001)
vB^2 = 1.29528 x 10^-3 m^2/s^2
Taking the square root of both sides:
vB = 0.036 J/s
Thus, the speed of the small object when it reaches point B is approximately 0.036 m/s.