Question

Two tiny sphere carrying a charges 1.8 µC and 2.8 µC are located at 40 cm apart. The potential at the mid-point of the line joining the two charges is

• A. 4.3 x 104 V
• B. 3.8 x 104 V
• C. 3.6 x 103 V
• D. 2.1 x 105 V

Answer 1

Answer: (D)

2.1 x 105 V

Answer 2

Let's consider the two charges, q1 = 1.8 µC and q2 = 2.8 µC, which are located 40 cm (0.4 m) apart. The midpoint between the charges is at a distance of 20 cm (0.2 m) from each charge.
The potential at the midpoint due to q1 is given by:
V1 = k * (q1 / r1)
Where r1 is the distance from q1 to the midpoint. In this case, r1 = 0.2 m.
Similarly, the potential at the midpoint due to q2 is given by:
V2 = k * (q2 / r2)
Where r2 is the distance from q2 to the midpoint. In this case, r2 = 0.2 m. The total potential at the midpoint is the sum of the potentials due to q1 and q2:
V = V1 + V2
Now let's calculate the potential using the given values:
V1 = (8.99 x 109 N m²/C²) * (1.8 x 10-6 C) / (0.2 m)
≈ 8.095 x 104 V
V2 = (8.99 x 109 N m²/C²) * (2.8 x 10-6 C) / (0.2 m)
≈ 1.257 x 105 V
V = V1 + V2
≈ 8.095 x 104 V + 1.257 x 105 V
≈ 2.0662 x 105 V
Therefore, the potential at the midpoint of the line joining the two charges is approximately 2.0662 x 105 V. Among the given options, the closest value is (D) 2.1 x 105 V.