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Electrical Potential Energy of Point Charges

Created by @nathanedwards
 at November 2nd 2023, 4:03:08 pm.
AP_Physics_2-Questions
#charges
#electric-potential-energy
#conducting-wire

Question:

Two identical point charges, each with a charge of +4 μC, are placed at a distance of 3 meters apart in a vacuum. The charges are then connected by a conducting wire and brought into contact with each other for a very short time. After they separate, the charges are placed back to their original positions.

(a) Calculate the initial electric potential energy of the system when the charges are initially placed at a distance of 3 meters.

(b) Determine the final electric potential energy of the system after the charges separate.

(c) Calculate the change in electric potential energy of the system as a result of bringing the charges in contact and then separating them.

(d) During the process of bringing the charges in contact and separating them, is any electrical energy converted to a different form? Explain your answer.

Answer:

(a) The electric potential energy of the system is given by the formula:

PE=k⋅q1⋅q2r PE = \frac{k \cdot q_1 \cdot q_2}{r}

where k is the electrostatic constant (k ≈ 8.99 × 10^9 N · m^2/C^2), q_1 and q_2 are the charges, and r is the distance between the charges.

Substituting the given values, the initial electric potential energy is:

PEinitial=(8.99×109 N⋅m2/C2)⋅(4×10−6 C)⋅(4×10−6 C)3 m=3.99 J PE_{initial} = \frac{(8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (4 \times 10^{-6} \, C) \cdot (4 \times 10^{-6} \, C)}{3 \, m} = 3.99 \, J

Therefore, the initial electric potential energy of the system is approximately 3.99 J.

(b) After the charges separate, the distance between them doubles to 6 meters. Using the same formula as before, the final electric potential energy is:

PEfinal=(8.99×109 N⋅m2/C2)⋅(4×10−6 C)⋅(4×10−6 C)6 m=1.33 J PE_{final} = \frac{(8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (4 \times 10^{-6} \, C) \cdot (4 \times 10^{-6} \, C)}{6 \, m} = 1.33 \, J

Therefore, the final electric potential energy of the system is approximately 1.33 J.

(c) The change in electric potential energy is given by:

ΔPE=PEfinal−PEinitial \Delta PE = PE_{final} - PE_{initial}

Substituting the given values, the change in electric potential energy is:

ΔPE=1.33 J−3.99 J=−2.66 J \Delta PE = 1.33 \, J - 3.99 \, J = -2.66 \, J

Therefore, the change in electric potential energy of the system is approximately -2.66 J.

(d) During the process of bringing the charges in contact and then separating them, electrical potential energy is converted into thermal energy. When the charges are brought into contact, some heat is generated due to the resistance in the conducting wire. This heat energy is dissipated into the surroundings and cannot be recovered. Therefore, electrical energy is converted into thermal energy, resulting in a decrease in the total electrical potential energy of the system.

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