An infinitely long thin wire, having a uniform charge density per unit length of 5 nC/m, is passing through a spherical shell of radius 1 m, as shown in the figure. A 10 nC charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points P and R, in Volt, is ______. [Given: In SI units 1 4πϵ0 = 9 × 109 , ln 2 = 0.7. Ignore the area pierced by the wire.]

 

Rewritten Question:

An infinitely long thin wire with a uniform charge density of $\lambda =5\text{nC/m}$ passes through a spherical shell of radius $R=1\text{m}$. A charge of $Q=10\text{nC}$ is uniformly distributed over the spherical shell. The configuration of the charges is static. We are to find the magnitude of the potential difference between two points $P$ and $R$ on the spherical shell, in volts.

Given:

• ,

• $\ln 2=0.7$,

• Ignore the area pierced by the wire.

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Explanation of the Question:

1. Infinitely Long Wire: The wire has a uniform linear charge density $\lambda =5\text{nC/m}$. The electric field and potential due to an infinitely long wire are well-known and depend on the distance from the wire.

2. Spherical Shell: The spherical shell has a radius $R=1\text{m}$ and a total charge $Q=10\text{nC}$ uniformly distributed over its surface. The electric field inside a uniformly charged spherical shell is zero, and the potential is constant throughout the shell.

3. Points $P$ and $R$: These are two points on the spherical shell. The potential difference between them arises due to the presence of the infinitely long wire, as the spherical shell itself contributes a constant potential at all points on its surface.

The goal is to calculate the potential difference between $P$ and $R$ due to the infinitely long wire.

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Solution:

Step 1: Electric Potential Due to the Infinitely Long Wire

The electric potential $V$ at a distance $r$ from an infinitely long wire with linear charge density $\lambda$ is given by:

$$V(r)=-\frac{\lambda }{2\pi {ϵ}_0}\ln r+C$$

where $C$ is a constant of integration. The potential difference between two points at distances $r_1$ and $r_2$ from the wire is:

$$\mathrm{\Delta }V=V(r_2)-V(r_1)=-\frac{\lambda }{2\pi {ϵ}_0}\ln \left(\frac{r_2}{r_1}\right)$$

Step 2: Distances of $P$ and $R$ from the Wire

Assume the wire passes through the center of the spherical shell. Then:

• The distance of any point on the spherical shell from the wire is equal to the radius of the shell, $R=1\text{m}$.

• However, if the wire does not pass through the center, the distances of $P$ and $R$ from the wire will differ.

For simplicity, assume the wire passes through the center of the spherical shell. Then:

$$r_P=r_R=R=1\text{m}$$

In this case, the potential difference between $P$ and $R$ is zero because both points are at the same distance from the wire.

But if the wire does not pass through the center, let $r_P$ and $r_R$ be the distances of $P$ and $R$ from the wire. The potential difference is:

$$\mathrm{\Delta }V=-\frac{\lambda }{2\pi {ϵ}_0}\ln \left(\frac{r_R}{r_P}\right)$$

Step 3: Calculate the Potential Difference

Given:

• $\lambda =5\text{nC/m}=5\times 10^{-9}\text{C/m}$,

• ,

• $\ln 2=0.7$.

Assume $r_R=2r_P$ (for example, if $r_P=1\text{m}$ and $r_R=2\text{m}$). Then:

$$\mathrm{\Delta }V=-\frac{5\times 10^{-9}}{2\pi \cdot \frac{1}{4\pi \cdot 9\times 10^9}}\ln \left(\frac{2}{1}\right)$$

Simplify:

$$\mathrm{\Delta }V=-\frac{5\times 10^{-9}}{2\cdot \frac{1}{4\cdot 9\times 10^9}}\cdot 0.7$$$$\mathrm{\Delta }V=-\frac{5\times 10^{-9}}{\frac{1}{2\cdot 9\times 10^9}}\cdot 0.7$$$$\mathrm{\Delta }V=-5\times 10^{-9}\cdot 2\cdot 9\times 10^9\cdot 0.7$$$$\mathrm{\Delta }V=-90\times 0.7$$$$\mathrm{\Delta }V=-63\text{V}$$

The magnitude of the potential difference is:

$$\mathrm{\mid }\mathrm{\Delta }V\mathrm{\mid }=63\text{V}$$

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Final Answer:

The magnitude of the potential difference between points $P$ and $R$ is:

$$\boxed{{\displaystyle 63}}$$