An infinitely long thin wire, having a uniform charge density per unit length of 5nC/m , is passing

Q: Why don't we calculate the absolute potential for the wire?

The absolute potential for an infinitely long wire is undefined (it diverges to infinity if we try to set the reference at infinity). We can only calculate the potential difference between two points at finite distances.

Q: Why is the potential at P due to the sphere calculated using the sphere's radius and not P's distance?

Point P is located inside the uniformly charged spherical shell. By Gauss's Law, the electric field inside is zero, meaning the potential is constant everywhere inside and equals the potential at the surface.

Q: How did $\ln(4)$ become $1.4$?

Using logarithm properties, $\ln(4) = \ln(2^2) = 2\ln(2)$. Since we are given $\ln(2) = 0.7$, $2 \times 0.7 = 1.4$.

Q: Does the wire piercing the shell affect the shell's charge distribution?

The problem explicitly states to 'ignore the area pierced by the wire' and that the 'configuration of the charges remains static'. This means we assume the charge distributions do not interact or redistribute.

Q: Can we just add the electric fields and then integrate?

Yes, you can! Since $V = -\int \vec{E} \cdot d\vec{r}$, the integral of a sum is the sum of the integrals. Calculating the potential differences separately and adding them (superposition) is mathematically identical and much easier.

Question

An infinitely long thin wire, having a uniform charge density per unit length of $5	ext{ nC/m}$ , is passing through a spherical shell of radius $1	ext{ m}$ , as shown in the figure. A $10	ext{ nC}$ 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 $\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9$ , $\ln 2 = 0.7$ . Ignore the area pierced by the wire.]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a complex electrostatic system. We have two distinct charge distributions occupying the same space: an infinitely long straight wire carrying a uniform linear charge density $\lambda$, and a spherical shell carrying a uniform surface charge $Q$. Both of these are perfectly centered at the origin. Our mission is to find the total potential difference between two specific points, P and R.

When faced with multiple charge distributions, the most powerful tool in our arsenal is the **Principle of Superposition**. It tells us that we don't need to panic about how the fields interact; we can simply calculate the potential difference due to the wire as if the sphere didn't exist, then calculate the potential difference due to the sphere as if the wire didn't exist, and finally add them together.

$$V_{	ext{total}} = V_{	ext{wire}} + V_{	ext{sphere}}$$

### The Master Equation for the Wire

Let's first tackle the infinitely long wire. The electric field of a line charge is radial and its magnitude is given by $E = \frac{2k\lambda}{r}$. To find the potential difference between two points at distances $r_R$ and $r_P$ from the wire, we integrate the electric field:

$$(V_P - V_R)_{	ext{wire}} = \int_{r_P}^{r_R} \frac{2k\lambda}{r} dr = 2k\lambda \ln\left(\frac{r_R}{r_P}ight)$$

Now, let's substitute the values given in the problem. The linear charge density $\lambda$ is $5 	imes 10^{-9} 	ext{ C/m}$. Point P is at a distance of $0.5 	ext{ m}$, and point R is at a distance of $2 	ext{ m}$ from the wire.

$$(V_P - V_R)_{	ext{wire}} = 2(9 	imes 10^9)(5 	imes 10^{-9}) \ln\left(\frac{2}{0.5}ight)$$

The powers of ten beautifully cancel out, leaving us with $90 \ln(4)$. Using the properties of logarithms, $\ln(4) = 2 \ln(2)$. Since we are given $\ln(2) = 0.7$, this becomes $90 	imes 1.4$, which equals **$126 	ext{ V}$**.

### The Master Equation for the Sphere

Next, we turn our attention to the spherical shell. A uniformly charged shell has a fascinating property: the electric field inside is zero. This means the potential everywhere inside the shell is constant and equal to the potential at its surface. Outside the shell, it behaves exactly like a point charge located at its center.

$$V_{	ext{inside}} = \frac{kQ}{R_{	ext{shell}}}, \quad V_{	ext{outside}} = \frac{kQ}{r}$$

Point P is located at $0.5 	ext{ m}$, which is inside the $1 	ext{ m}$ radius shell. Therefore, its potential is determined by the shell's radius. Point R is at $2 	ext{ m}$, which is outside the shell. Let's set up the potential difference:

$$(V_P - V_R)_{	ext{sphere}} = \frac{kQ}{1} - \frac{kQ}{2} = kQ \left( 1 - \frac{1}{2} ight)$$

Substituting the charge $Q = 10 	imes 10^{-9} 	ext{ C}$, we get:

$$(V_P - V_R)_{	ext{sphere}} = (9 	imes 10^9)(10 	imes 10^{-9}) \left( \frac{1}{2} ight) = 90 	imes \frac{1}{2} = \mathbf{45 	ext{ V}}$$

### Final Calculation

We have successfully isolated and conquered both parts of the problem. Now, we bring them together using superposition. We simply add the potential differences from both the wire and the sphere to find the total potential difference.

$$V_P - V_R = 126 	ext{ V} + 45 	ext{ V} = \mathbf{171 	ext{ V}}$$

And there we have it! By breaking down a complex configuration into fundamental, manageable pieces, we arrived at the elegant final answer of 171 Volts.

Enter Numerical Value:

Visualized Solution

$V = V_{wire} + V_{sphere}$

$(V_{P} - V_{R})_{wire} = 2 k λ ln (\frac{r _{R}}{r _{P}})$

$(V_{P} - V_{R})_{wire} = 2 (9 \times 1 0^{9}) (5 \times 1 0^{- 9}) ln (\frac{2}{0.5})$

$(V_{P} - V_{R})_{wire} = 126 V$

$V_{inside} = \frac{k Q}{R _{shell}}, V_{outside} = \frac{k Q}{r}$

$(V_{P} - V_{R})_{sphere} = k Q (\frac{1}{1} - \frac{1}{2})$

$(V_{P} - V_{R})_{sphere} = 45 V$

V_P - V_R = 171 \text{ V}

\text{Superposition Principle: } V_{\text{total}} = \sum V_i

The Sigma Insight: Electric Potential and Potential Difference

Analyzing the Setup

The Master Equation for the Wire

The Master Equation for the Sphere

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=Vwire​+Vsphere​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (VP​−VR​)wire​=2kλln(rP​rR​​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (VP​−VR​)wire​=2(9×109)(5×10−9)ln(0.52​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (VP​−VR​)wire​=126 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vinside​=Rshell​kQ​,Voutside​=rkQ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (VP​−VR​)sphere​=kQ(11​−21​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (VP​−VR​)sphere​=45 V