Two particles each with charge +2.68 µC are located on the x axis. One is at x = 1.00 m, and the other is at x = −1.00 m.
(a) Determine the electric potential on the y axis at y = 0.660 m.
(b) Calculate the change in electric potential energy of the system as a third charged particle of -3.66 µC is brought from infinitely far away to a position on the y axis at y = 0.660 m.
(a) Well, well, well, looks like we have two charged particles having a little party on the x axis. One is at x = 1.00 m, and the other is at x = -1.00 m. You want to know the electric potential on the y axis at y = 0.660 m? Alright, let's get to it.
First, let's determine the distance of each particle from the point (0,0.660 m) on the y axis. The distance from the particle at x = 1.00 m would be D1 = √(1.00 m)^2 + (0.660 m)^2. And the distance from the particle at x = -1.00 m would be D2 = √(-1.00 m)^2 + (0.660 m)^2. Let's calculate those.
D1 = √(1.00 m)^2 + (0.660 m)^2 = 1.229 m
D2 = √(-1.00 m)^2 + (0.660 m)^2 = 1.229 m
Alright, alright, the distances are equal, it's a tie! Now, let's calculate the electric potential V for each particle using the formula V = kQ / r, where k is the Coulomb's constant (k = 8.99 x 10^9 Nm^2/C^2), Q is the charge of the particle, and r is the distance from the particle to the point (0,0.660 m).
For the particle at x = 1.00 m:
V1 = (8.99 x 10^9 Nm^2/C^2)(2.68 x 10^-6 C) / 1.229 m = 1.89 x 10^6 V
And for the particle at x = -1.00 m:
V2 = (8.99 x 10^9 Nm^2/C^2)(2.68 x 10^-6 C) / 1.229 m = 1.89 x 10^6 V
Since the distances were equal, the electric potential at y = 0.660 m due to each particle is the same. Therefore, the total electric potential at y = 0.660 m is the sum of the individual electric potentials.
V_total = V1 + V2 = 1.89 x 10^6 V + 1.89 x 10^6 V = 3.78 x 10^6 V
So, the electric potential on the y axis at y = 0.660 m is 3.78 x 10^6 V. Keep that voltage in check!
(b) Ahh, now you want to calculate the change in electric potential energy of the system as a third charged particle joins the party. This third particle has a charge of -3.66 µC and is brought from infinitely far away to a position on the y axis at y = 0.660 m. Let's see what happens.
The change in electric potential energy ΔU is given by ΔU = qΔV, where q is the charge of the third particle and ΔV is the change in electric potential.
First, let's calculate the change in electric potential ΔV for the third particle. Since it's brought in from infinitely far away, the electric potential at infinity is considered to be zero. So, ΔV = V_total - V_infinity.
ΔV = 3.78 x 10^6 V - 0 V = 3.78 x 10^6 V
Now, we can calculate the change in electric potential energy.
ΔU = qΔV = (-3.66 x 10^-6 C)(3.78 x 10^6 V) = -13.86 J
So, the change in electric potential energy of the system as the third charged particle joins the party is -13.86 J. Looks like things got a bit negative there!
To solve this problem, we'll need to use the formula for electric potential due to a point charge, as well as the formula for electric potential energy.
(a) To find the electric potential at a point on the y axis, we'll need to calculate the contributions from both charges using the formula:
V = k * (q1/r1 + q2/r2)
Where:
- V is the electric potential at the point
- k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2)
- q1 and q2 are the charges of the particles
- r1 and r2 are the distances between the particles and the point
Let's substitute the given values into the formula:
V = (8.99 x 10^9 Nm^2/C^2) * [(2.68 x 10^-6 C) / (0.660 m - 1.00 m) + (2.68 x 10^-6 C) / (0.660 m + 1.00 m)]
Simplifying the equation, we get:
V = (8.99 x 10^9 Nm^2/C^2) * [2(2.68 x 10^-6 C) / 1.66 m]
By calculating the numbers, we find:
V ≈ 2.74 x 10^6 Volts
So, the electric potential on the y-axis at y = 0.660 m is approximately 2.74 x 10^6 Volts.
(b) To calculate the change in electric potential energy, we need to consider the charge of the third particle and its distance from the other two charges. The formula for electric potential energy between two charges is:
ΔPE = k * ((q1 * q3) / r13 - (q2 * q3) / r23)
Where:
- ΔPE is the change in electric potential energy
- q3 is the charge of the third particle
- r13 is the distance between the first particle and the third particle
- r23 is the distance between the second particle and the third particle
For our case, since the third particle is brought from infinitely far away, the initial electric potential energy is zero. So, we can calculate the change in electric potential energy by only considering the final position of the third charge:
ΔPE = k * (q1 * q3 / r13 - q2 * q3 / r23)
Substituting the given values:
ΔPE = (8.99 x 10^9 Nm^2/C^2) * [(2.68 x 10^-6 C) * (-3.66 x 10^-6 C) / (1.00 m) - (2.68 x 10^-6 C) * (-3.66 x 10^-6 C) / (1.00 m)]
Simplifying the equation, we get:
ΔPE = (8.99 x 10^9 Nm^2/C^2) * [(-2.68 x 3.66) x 10^-6 C^2 / (1.00 m)]
By calculating the numbers, we find:
ΔPE ≈ -3.89 x 10^-3 Joules
So, the change in electric potential energy of the system as the third charged particle is brought from infinitely far away to a position on the y-axis at y = 0.660 m is approximately -3.89 x 10^-3 Joules.