Question

At the centre of a half ring of radius $R = 10 \, \text{cm}$ and linear charge density $4n \, \text{C m}^{-1}$, the potential is $x \pi \, \text{V}$. The value of $x$ is ______.

Answer 1

Correct Answer: 36

To find the potential at the center of a half ring of radius \( R = 10 \, \text{cm} \) with a linear charge density \( \lambda = 4 \, \text{nC} \, \text{m}^{-1} \), we use the formula for the electric potential due to a charged arc: \( V = \frac{k \lambda L}{R} \), where \( k \) is Coulomb's constant \( \left( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \right) \), \( L \) is the arc length of the half ring, and \( R \) is the radius. The arc length \( L \) for a half ring is \( \pi R \).

First, convert the given linear charge density to coulombs per meter: \( \lambda = 4 \times 10^{-9} \, \text{C/m} \). The radius is \( R = 0.1 \, \text{m} \).

The arc length \( L = \pi R = \pi \times 0.1 \, \text{m} = 0.1\pi \, \text{m} \).

Substituting the values into the potential formula, we have:

\( V = \frac{(8.99 \times 10^9) \times (4 \times 10^{-9}) \times (0.1\pi)}{0.1} \).

Simplifying, \( V = 8.99 \times 4 \times \pi \).

Calculating gives \( V = 35.96\pi \, \text{V} \).

We identify \( V = x\pi \) and solve for \( x \), yielding \( x = 35.96 \).

Round \( x \) to the nearest whole number: \( x = 36 \).

This solution fits within the expected range of 36,36.

Answer 2

The potential at the center of a half-ring is given by:

\( V = \frac{KQ}{R} \)

where:

• \( Q = \lambda \pi R \)

Substituting:

\( V = \frac{K \lambda \pi R}{R} \)
\( V = K \lambda \pi \)

Given:

• \( K = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \),
• \( \lambda = 4 \times 10^{-9} \, \text{C/m} \)

\( V = 9 \times 10^9 \cdot 4 \times 10^{-9} \cdot \pi \) 
\( V = 36\pi \, \text{V} \)

Thus, \( x = 36 \).

Final Answer: \( x = 36 \).