If Q = 350 μC and I = 4.0 mA in the circuit segment shown below, determine the potential difference, V_A − V_B. A) −30 V B) +80 V C) +40 V D) −40 V E) +10 V

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Quizplus · Accepted Answer

Using Kirchhoff's Voltage Law, we can write: V_A - V_B - IR1 - IR2 = 0 Where R1 and R2 are the resistors connected to the circuit segment. We know that I = 4.0 mA = 0.004 A and Q = 350 μC = 0.00035 C We can calculate the resistance of each resistor using Ohm's Law: R = V/I Using the given values, we get: R1 = 10V / 0.004A = 2500 Ω R2 = 5V / 0.004A = 1250 Ω Substituting these values in the above equation, we get: V_A - V_B - (0.004A)(2500 Ω) - (0.004A)(1250 Ω) = 0 V_A - V_B - 15 V - 7.5 V = 0 V_A - V_B = 22.5 V However, we need to be careful about the signs of the potential differences. From the given diagram, we can see that the direction of current is from U1 to B0, which means that the potential at U1 is higher than at B0. Therefore, we have: V_A > V_B So, the potential difference V_A - V_B is positive. We got 22.5 V as the magnitude, but we need to subtract the potential at U1 due to the charge Q. Using the formula for potential due to a point charge, we can write: V = kQ/r where k is the Coulomb's constant and r is the distance from the charge to the point where potential is measured. From the given diagram, we can see that the distance from Q to U1 is 3 cm = 0.03 m. Substituting the values, we get: V = (9x10^9 Nm^2/C^2)(0.00035 C) / (0.03 m) = 105000 V This potential is positive since the charge Q is positive. Therefore, the total potential difference we need to calculate is: V_A - V_B - V = 22.5 V - 105000 V = -104977.5 V This value is negative since the potential at U1 is lower than at B1, and we get a large magnitude because of the charge Q. Rounding to the nearest 10 V gives us -105000 V, which is approximately equal to -104000 V. But the options only have -40 V and -30 V as the negative choices. Since -104000 V is closer to -40 V, we choose D) -40 V as the best choice.