FIGURE 20-10
-Two point charges of magnitude +4.00 μC and +2.00 μC are placed at the opposite corners of a rectangle as shown in Figure 20-10. What is the potential at point B due to these two charges?

Question

FIGURE 20-10 
-Two point charges of magnitude +4.00 μC and +2.00 μC are placed at the opposite corners of a rectangle as shown in Figure 20-10. What is the potential at point B due to these two charges?

Quizplus · Accepted Answer

The potential due to point charges can be calculated using the formula:

V = kq/r

where V is the potential, k is the Coulomb constant, q is the charge, and r is the distance between the point charge and the point where potential is to be calculated.

For point B, the potential due to the +4.00 μC charge is:

V1 = kq1/r1 = (9 × 10^9 N m^2/C^2)(4.00 × 10^-6 C)/(1.00 m) = 3.60 × 10^3 V

For point B, the potential due to the +2.00 μC charge is:

V2 = kq2/r2 = (9 × 10^9 N m^2/C^2)(2.00 × 10^-6 C)/(√2 m) = 1.27 × 10^3 V

The total potential at point B is the sum of the potentials due to the two charges:

V = V1 + V2 = 3.60 × 10^3 V + 1.27 × 10^3 V = 4.87 × 10^3 V

Converting to scientific notation, we get:

V = 4.87 × 10^3 V = 4.87 × 10^3 × 10^-6 kV = 4.87 × 10^-3 kV = 4.87 mV

Finally, since the question asks for the potential in volts, we need to convert millivolts to volts:

V = 4.87 mV = 4.87 × 10^-3 V

The closest answer choice to this value is E) 113 × 10^-3 V, which is equivalent to 113 mV, so the answer is E).

FIGURE 20-10 -Two Point Charges of Magnitude +4.00 μC and +2.00 μC