Question:
Consider a system where two point charges are positioned as follows:
(a) What is the electric potential at point P?
(b) Is the electric potential at point P positive or negative? Explain your reasoning.
(c) If a test charge of +2 μC is placed at point P, what is the electric potential energy of the test charge?
(d) If the test charge from part (c) is moved to a point Q at coordinates (0, 4), what is the change in electric potential energy? Is the work done by the electric field positive or negative? Explain your reasoning.
Answer:
(a) To find the electric potential at point P, we need to calculate the contributions of each charge to the overall potential.
Using the equation for electric potential due to a point charge:
V = kQ/r,
where V is the electric potential, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the charge.
For charge Q1 at the origin (0, 0): V1 = (9 x 10^9 Nm^2/C^2) * (4 x 10^-6 C) / (3m)
For charge Q2 at point P (3, 0): V2 = (9 x 10^9 Nm^2/C^2) * (-6 x 10^-6 C) / (3m)
The total electric potential at point P is the sum of the individual potentials: V_total = V1 + V2
Substituting in the values: V_total = [(9 x 10^9 Nm^2/C^2) * (4 x 10^-6 C) / (3m)] + [(9 x 10^9 Nm^2/C^2) * (-6 x 10^-6 C) / (3m)]
Simplifying the expression gives: V_total = 0 V
Therefore, the electric potential at point P is 0 V.
(b) The electric potential at point P is 0 V, which means there is no net influence on a positive test charge placed at that point. Since potential is a scalar quantity and can be positive, negative, or zero, in this case, the electric potential at point P is neither positive nor negative.
(c) To calculate the electric potential energy (U) of the test charge placed at point P, we'll use the equation:
U = q * V,
where U is the electric potential energy, q is the charge of the test charge, and V is the electric potential at P.
Given q = +2 μC and V = 0 V (as determined in part a), we can substitute these values into the equation:
U = (+2 x 10^-6 C) * (0 V)
U = 0 J
Therefore, the electric potential energy of the test charge at point P is 0 J.
(d) To find the change in electric potential energy, we'll use the equation:
ΔU = U_final - U_initial,
where ΔU is the change in electric potential energy, U_final is the electric potential energy at point Q, and U_initial is the electric potential energy at point P.
Given that U_initial = 0 J (as calculated in part c), we need to find the U_final at point Q.
Using the same equation U = q * V, where V is the electric potential at point Q, we can find U_final:
V_Q1 = (9 x 10^9 Nm^2/C^2) * (4 x 10^-6 C) / (5m)
V_Q2 = (9 x 10^9 Nm^2/C^2) * (-6 x 10^-6 C) / (4m)
V_total_Q = V_Q1 + V_Q2
Substituting the values and simplifying, we get: V_total_Q = 8 V
Now, substitute the value of U_initial and U_final into the equation:
ΔU = 8 V * (+2 x 10^-6 C) - 0 J
ΔU = 16 x 10^-6 J
Therefore, the change in electric potential energy is 16 x 10^-6 J.
Since the charge is moving against the electric field from point P to point Q, work is done on the charge. Therefore, the work done by the electric field is negative.