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

The Correct Option is D

Given two charges \( q_1 = 1.8 \, \mu\text{C} \) and \( q_2 = 2.8 \, \mu\text{C} \), located 40 cm (0.4 m) apart, calculate the electric potential at the midpoint between the charges. The midpoint is at a distance of 20 cm (0.2 m) from each charge.

Step 1: Formula for Electric Potential

The electric potential \( V \) at a point due to a charge \( q \) is given by the formula:

\( V = k \frac{q}{r} \)

Where:

• k: Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \)
• q: The charge producing the potential
• r: The distance from the charge to the point where the potential is being calculated

Step 2: Calculate the potential at the midpoint due to each charge:

Potential due to \( q_1 \) (V1):

\( V_1 = k \frac{q_1}{r_1} \)

Where \( r_1 = 0.2 \, \text{m} \) is the distance from \( q_1 \) to the midpoint. Substituting the given values:

\( V_1 = (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \times \frac{1.8 \times 10^{-6} \, \text{C}}{0.2 \, \text{m}} \approx 8.095 \times 10^4 \, \text{V} \)

Potential due to \( q_2 \) (V2):

\( V_2 = k \frac{q_2}{r_2} \)

Where \( r_2 = 0.2 \, \text{m} \) is the distance from \( q_2 \) to the midpoint. Substituting the given values:

\( V_2 = (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \times \frac{2.8 \times 10^{-6} \, \text{C}}{0.2 \, \text{m}} \approx 1.257 \times 10^5 \, \text{V} \)

Step 3: Calculate the total potential at the midpoint:

The total potential is the sum of the potentials due to \( q_1 \) and \( q_2 \):

\( V = V_1 + V_2 \)

Substituting the values for \( V_1 \) and \( V_2 \):

\( V = 8.095 \times 10^4 \, \text{V} + 1.257 \times 10^5 \, \text{V} \approx 2.0662 \times 10^5 \, \text{V} \)

Therefore, the potential at the midpoint between the two charges is approximately: \( 2.0662 \times 10^5 \, \text{V} \). The closest value is (D) \( 2.1 \times 10^5 \, \text{V} \).

Answer 2

Given:

• q₁ = 1.8 μC = 1.8 × 10⁻⁶ C 
• q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

Distance between the two spheres is: 40 cm = 0.4 m

For the midpoint, \( r_1 = r_2 = \frac{0.40}{2} = 0.2 \, \text{m} \)

Potential at point O:

\( V = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_1}{r_1} + \frac{q_2}{r_2} \right] = \frac{9 \times 10^9}{0.2} \times (1.8 \times 10^{-6} + 2.8 \times 10^{-6}) \)

Calculating the potential:

\( V = 2.1 \times 10^5 \, \text{V} \)

Therefore, the potential at point O is \( 2.1 \times 10^5 \, \text{V} \).

Answer 3

We are given the following information: 

• Charge on the first sphere, \( q_1 = 1.8 \, \mu\text{C} = 1.8 \times 10^{-6} \, \text{C} \)
• Charge on the second sphere, \( q_2 = 2.8 \, \mu\text{C} = 2.8 \times 10^{-6} \, \text{C} \)
• Distance between the spheres, \( d = 40 \, \text{cm} = 0.40 \, \text{m} \)
• Coulomb's constant, \( k \approx 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \)

We need to find the electric potential at the midpoint of the line joining the two charges.

The midpoint is located at a distance \( r \) from each charge, where:

\[ r = \frac{d}{2} = \frac{0.40 \, \text{m}}{2} = 0.20 \, \text{m} \]

The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by:

\[ V = k \frac{q}{r} \]

Electric potential is a scalar quantity. The total potential at the midpoint is the algebraic sum of the potentials due to each individual charge at that point.

Potential due to \( q_1 \) at the midpoint:

\[ V_1 = k \frac{q_1}{r} \]

Potential due to \( q_2 \) at the midpoint:

\[ V_2 = k \frac{q_2}{r} \]

Total potential at the midpoint, \( V_{mid} \):

\[ V_{mid} = V_1 + V_2 = k \frac{q_1}{r} + k \frac{q_2}{r} \]

\[ V_{mid} = \frac{k}{r} (q_1 + q_2) \]

First, calculate the sum of the charges:

\[ q_1 + q_2 = (1.8 \times 10^{-6}) + (2.8 \times 10^{-6}) = (1.8 + 2.8) \times 10^{-6} \]

\[ q_1 + q_2 = 4.6 \times 10^{-6} \, \text{C} \]

Now, substitute the values into the formula for \( V_{mid} \):

\[ V_{mid} = \frac{9 \times 10^9 \, \text{N m}^2 \text{C}^{-2}}{0.20 \, \text{m}} \times (4.6 \times 10^{-6} \, \text{C}) \]

\[ V_{mid} = \frac{9}{0.20} \times 4.6 \times 10^{9-6} \, \text{V} \]

\[ V_{mid} = 45 \times 4.6 \times 10^3 \, \text{V} \]

\[ V_{mid} = 207 \times 10^3 \, \text{V} \]

\[ V_{mid} = 2.07 \times 10^2 \times 10^3 \, \text{V} \]

\[ V_{mid} = 2.07 \times 10^5 \, \text{V} \]

This value is approximately \( 2.1 \times 10^5 \, \text{V} \).