Let E1(r), E2(r) and E3(r) be the respective electric fields at a distance r from a point charge Q,

Q: Why is the electric field of the infinite plane constant?

For an infinite plane, the electric field lines are parallel and uniformly spaced. As you move away from the plane, the field lines don't spread out, so the field strength remains constant.

Q: How do we know the electric field of a line charge is inversely proportional to $r$?

By applying Gauss's Law to a cylindrical Gaussian surface around the line charge, the surface area grows linearly with $r$ ($2\pi r L$), leading to an electric field that falls off as $1/r$.

Q: Why does the point charge field fall off as $1/r^2$?

The electric field lines from a point charge spread out in three dimensions. The area of a spherical surface grows as $r^2$ ($4\pi r^2$), so the field strength decreases as $1/r^2$.

Q: Can we use Gauss's Law for all three distributions?

Yes! Gauss's Law is perfectly suited for these highly symmetric charge distributions: spherical symmetry for the point charge, cylindrical symmetry for the line charge, and planar symmetry for the sheet.

Q: What if the distance was doubled instead of halved?

If the distance was doubled to $2r_0$, $E_1$ would become $1/4$ of its original value, $E_2$ would become $1/2$, and $E_3$ would remain the same.

Question

Let $E_1(r)$, $E_2(r)$ and $E_3(r)$ be the respective electric fields at a distance $r$ from a point charge $Q$, an infinitely long wire with constant linear charge density $\lambda$, and an infinite plane with uniform surface charge density $\sigma$. If $E_1(r_0) = E_2(r_0) = E_3(r_0)$ at a given distance $r_0$, then

Options:

(A) $Q = 4\sigma\pi r_0^2$

(B) $r_0 = \frac{\lambda}{2\pi\sigma}$

(C) $E_1(r_0/2) = 2E_2(r_0/2)$

(D) $E_2(r_0/2) = 4E_3(r_0/2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem presents us with three distinct charge distributions: a point charge $Q$, an infinitely long wire with linear charge density $\lambda$, and an infinite plane with surface charge density $\sigma$. We are given that at a specific distance $r_0$, the electric fields produced by all three distributions are perfectly equal.

To begin, we must recall the fundamental formulas for the electric field produced by each of these distributions at a distance $r$:
1. **Point Charge:** $E_1(r) = \frac{Q}{4\pi\varepsilon_0 r^2}$
2. **Line Charge:** $E_2(r) = \frac{\lambda}{2\pi\varepsilon_0 r}$
3. **Plane Charge:** $E_3(r) = \frac{\sigma}{2\varepsilon_0}$

Notice how each field depends differently on the distance $r$. The point charge field follows an inverse-square law ($1/r^2$), the line charge field follows an inverse law ($1/r$), and the plane charge field is completely independent of distance!

### Checking the First Two Options

We are given that $E_1(r_0) = E_2(r_0) = E_3(r_0)$. Let's use these equalities to test the first two options.

First, let's equate the field of the point charge to the field of the plane charge:
$$ \frac{Q}{4\pi\varepsilon_0 r_0^2} = \frac{\sigma}{2\varepsilon_0} $$

By rearranging the terms and solving for $Q$, we find:
$$ Q = 2\pi\sigma r_0^2 $$

Option (a) claims that $Q = 4\sigma\pi r_0^2$, which is clearly incorrect based on our derivation.

Next, let's equate the field of the line charge to the field of the plane charge:
$$ \frac{\lambda}{2\pi\varepsilon_0 r_0} = \frac{\sigma}{2\varepsilon_0} $$

Solving for $r_0$, we get:
$$ r_0 = \frac{\lambda}{\pi\sigma} $$

Option (b) states that $r_0 = \frac{\lambda}{2\pi\sigma}$, which is also incorrect.

### The Effect of Halving the Distance

Now, the problem asks us to evaluate the electric fields at half the original distance, $r_0/2$. This is where the different distance dependencies become crucial.

For the point charge, halving the distance increases the field by a factor of $2^2 = 4$:
$$ E_1(r_0/2) = \frac{Q}{4\pi\varepsilon_0 (r_0/2)^2} = 4E_1(r_0) $$

For the line charge, halving the distance increases the field by a factor of $2$:
$$ E_2(r_0/2) = \frac{\lambda}{2\pi\varepsilon_0 (r_0/2)} = 2E_2(r_0) $$

We know from the initial conditions that $E_1(r_0) = E_2(r_0)$. Therefore, we can substitute $E_2(r_0)$ into our new equation for $E_1$:
$$ E_1(r_0/2) = 4E_2(r_0) = 2(2E_2(r_0)) = 2E_2(r_0/2) $$

This perfectly matches option (c)! The new point charge field is exactly twice the new line charge field.

### Final Verification

Just to be thorough, let's check option (d). The plane charge field is uniform, meaning it doesn't change regardless of the distance:
$$ E_3(r_0/2) = E_3(r_0) $$

We already established that $E_2(r_0/2) = 2E_2(r_0)$. Since $E_2(r_0) = E_3(r_0)$, we can write:
$$ E_2(r_0/2) = 2E_3(r_0) = 2E_3(r_0/2) $$

Option (d) claims that $E_2(r_0/2) = 4E_3(r_0/2)$, which is incorrect. The correct relationship is a factor of 2, not 4.

**Conclusion:** The only correct statement is option (c).

Select Answer:

Visualized Solution

$E_{1}, E_{2}, E_{3} at r_{0}$

$Checking Option (a)$

$Checking Option (b)$

$Checking Option (c)$

$Checking Option (d)$

The Sigma Insight: Electric Field

Analyzing the Setup

Checking the First Two Options

The Effect of Halving the Distance

Final Verification

Similar Questions

Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q$ . The distance of the point P from the centre of the rod is $a = \frac{3}{2} L$ .

Charges $Q_{1}$ and $Q_{2}$ are at points $A$ and $B$ of a right angle triangle $O A B$ (see figure). The resultant electric field at point $O$ is perpendicular to the hypotenuse, then $Q_{1} / Q_{2}$ is proportional to

A cube of side $a$ has point charges $+ Q$ located at each of its vertices except at the origin, where the charge is $- Q$ . The electric field at the centre of cube is

Four point charges $- q, + q, + q$ and $- q$ are placed on Y-axis at $y = - 2 d$ , $y = - d$ , $y = + d$ and $y = + 2 d$ , respectively. The magnitude of the electric field $E$ at a point on the X-axis at $x = D$ , with $D >> d$ , will behave as

A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field $E$ at the centre $O$ is

Two point charges $q_{1} (10 μ C)$ and $q_{2} (- 25 μ C)$ are placed on the x-axis at $x = 1 m$ and $x = 4 m$ , respectively. The electric field (in V/m) at a point $y = 3 m$ on Y-axis is $(Take, \frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9} N-m^{2} C^{- 2})$

Three charged particles $A, B$ and $C$ with charges $- 4 q, 2 q$ and $- 2 q$ are present on the circumference of a circle of radius $d$ . The charged particles $A, C$ and centre $O$ of the circle formed an equilateral triangle as shown in figure. Electric field at $O$ along x-direction is

What will be the magnitude of electric field at point $O$ as shown in figure? Each side of the figure is $l$ and perpendicular to each other.

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​,E2​,E3​ at r0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (a)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (b)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (c)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (d)