A non-conducting ring of radius 0.5 m carries a total charge of 1.11 * 10^-10 C distributed non-unif

Q: Does the non-uniform charge distribution affect the electric field at the center?

Yes, the electric field at the center will generally be non-zero and its direction will depend on the exact charge distribution. However, the electric *potential* is a scalar, and since all charge elements are equidistant from the center, the potential only depends on the total charge.

Q: Why do we take the potential at infinity to be zero?

It's a standard convention in electrostatics for localized charge distributions. It provides a convenient reference point, making the potential at any point equal to the work done per unit charge to bring a test charge from infinity to that point.

Q: What if the integral was from the center to infinity instead?

The integral $\int_{0}^{\infty} - \mathbf{E} \cdot d\mathbf{l}$ would represent $V_{\infty} - V_{\text{centre}}$, which would be $0 - 2 = -2 \text{ V}$. The limits of integration determine the sign.

Q: Can we use Gauss's Law to solve this?

Gauss's Law is useful for finding the electric field for highly symmetric charge distributions. Here, the charge is non-uniform, so Gauss's Law isn't helpful for finding the field. Fortunately, we only need the potential, which can be found directly.

Q: Why is the answer exactly 2 when the calculation gives 1.998?

The value $1.11 \times 10^{-10}$ is likely an approximation for $\frac{1}{9} \times 10^{-9}$. If you use $q = \frac{10^{-9}}{9}$, the calculation becomes exactly $\frac{9 \times 10^9 \times 10^{-9} / 9}{0.5} = \frac{1}{0.5} = 2$.

Question

A non-conducting ring of radius $0.5	ext{ m}$ carries a total charge of $1.11 	imes 10^{-10}	ext{ C}$ distributed non-uniformly on its circumference producing an electric field $\mathbf{E}$ everywhere in space. The value of the integral $\int_{l=\infty}^{l=0} - \mathbf{E} \cdot d\mathbf{l}$ ($l=0$ being centre of the ring) in volt is

Options:

(A) $+2$

(B) $-1$

(C) $-2$

(D) $	ext{zero}$

Sigma Percentile · Accepted Answer

The problem asks us to evaluate a seemingly complex line integral of the electric field produced by a non-uniformly charged ring. At first glance, the non-uniformity of the charge distribution and the integral itself might look intimidating. However, the beauty of physics lies in recognizing the underlying concepts that simplify such problems.

### Decoding the Integral

Let's start by looking at the integral we need to evaluate:
$$ \int_{l=\infty}^{l=0} - \mathbf{E} \cdot d\mathbf{l} $$

We know from electrostatics that the electric field $\mathbf{E}$ is the negative gradient of the electric potential $V$. Mathematically, this is expressed as:
$$ dV = - \mathbf{E} \cdot d\mathbf{l} $$

Therefore, the line integral of the electric field from infinity to the center of the ring ($l=0$) is simply the potential difference between these two points:
$$ \int_{\infty}^{0} dV = V_{	ext{centre}} - V_{\infty} $$

By convention, the electric potential at infinity is taken to be zero ($V_{\infty} = 0$). Thus, the entire integral simplifies to just the electric potential at the center of the ring:
$$ \int_{l=\infty}^{l=0} - \mathbf{E} \cdot d\mathbf{l} = V_{	ext{centre}} $$

### The Magic of the Center

Now, we need to find the electric potential at the center of the ring. The problem states that the charge $q = 1.11 	imes 10^{-10} 	ext{ C}$ is distributed **non-uniformly** over the ring. Does this non-uniformity complicate our calculation?

Not at all! The electric potential is a scalar quantity. The potential $dV$ at the center due to a tiny charge element $dq$ on the ring is:
$$ dV = \frac{1}{4\pi\varepsilon_0} \frac{dq}{R} $$

Since every point on the ring is at the exact same distance $R$ from the center, $R$ is a constant for all charge elements. When we integrate to find the total potential, we can pull $R$ out of the integral:
$$ V_{	ext{centre}} = \int \frac{1}{4\pi\varepsilon_0} \frac{dq}{R} = \frac{1}{4\pi\varepsilon_0 R} \int dq $$

The integral of $dq$ over the entire ring is simply the total charge $q$. Therefore, the potential at the center is:
$$ V_{	ext{centre}} = \frac{q}{4\pi\varepsilon_0 R} $$

This is a profound result: **the potential at the center of a charged ring depends only on the total charge and the radius, regardless of how the charge is distributed!**

### Final Calculation

Now, it's just a matter of plugging in the given values:
*   Total charge, $q = 1.11 	imes 10^{-10} 	ext{ C}$
*   Radius, $R = 0.5 	ext{ m}$
*   Coulomb's constant, $\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$

Substituting these into our formula:
$$ V_{	ext{centre}} = \frac{(9 	imes 10^9) 	imes (1.11 	imes 10^{-10})}{0.5} $$

$$ V_{	ext{centre}} = \frac{9 	imes 1.11 	imes 10^{-1}}{0.5} $$

$$ V_{	ext{centre}} = \frac{0.999}{0.5} $$

Since $0.999$ is approximately $1$, we have:
$$ V_{	ext{centre}} \approx \frac{1}{0.5} = 2 	ext{ V} $$

Thus, the value of the integral is $+2 	ext{ V}$.

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Visualized Solution

$Relation between Electric Field and Potential$

$Potential Difference$

$Potential at the Centre of a Ring$

$Substitution$

$Final Calculation$

The Sigma Insight: Electric Potential and Potential Difference

Decoding the Integral

The Magic of the Center

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relation between Electric Field and Potential

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential at the Centre of a Ring

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation