An electric dipole is placed on X-axis in proximity to a line charge of linear charge density 3.0 x

Q: Why do we subtract the forces $F_1$ and $F_2$?

Because the electric field points radially outward. It pushes the positive charge away (repulsion) and pulls the negative charge in (attraction). Since they act in opposite directions, the net force is their difference.

Q: Why can't we use the standard dipole force formula $F = p \frac{dE}{dr}$?

That formula is an approximation valid only when the dipole length is extremely small compared to the distance from the source. Here, the dipole length (2 mm) is significant compared to its distance from the origin (10 mm), so we must calculate the exact forces.

Q: What is the direction of the net force?

The net force is repulsive, pointing away from the line charge along the positive X-axis. This is because the positive charge is closer to the line charge and experiences a stronger force than the negative charge.

Q: Why did we convert millimeters to meters?

To ensure dimensional consistency. The force is given in Newtons (SI unit) and the charge density in C/m (SI unit). Using millimeters would result in an incorrect order of magnitude for the final charge.

Q: Does the line charge exert a torque on the dipole in this setup?

No. Since both charges of the dipole lie exactly on the X-axis (along the electric field lines), the forces are collinear. There is no perpendicular distance between the lines of action of the forces, hence zero torque.

Question

An electric dipole is placed on X-axis in proximity to a line charge of linear charge density $3.0 	imes 10^{-6}$ C/m. Line charge is placed on Z-axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin, respectively. If total force of 4 N is exerted on the dipole, find out the amount of positive or negative charge of the dipole.

Options:

(A) 815.1 nC

(B) 8.8 $\mu$C

(C) 0.485 nC

(D) 4.44 $\mu$C

Sigma Percentile · Accepted Answer

The interaction between an electric dipole and a non-uniform electric field is a classic concept in electrostatics. In this problem, we are exploring the force exerted by an infinitely long line charge on a dipole placed nearby. Let's break down the physics and the math step-by-step!

### Analyzing the Setup

Imagine an infinitely long line charge placed along the Z-axis. This line charge has a linear charge density of $\lambda = 3.0 	imes 10^{-6} 	ext{ C/m}$.

Now, we place an electric dipole on the X-axis. The problem states that the positive charge $+q$ is at a distance of $r_1 = 10 	ext{ mm}$ from the origin, and the negative charge $-q$ is at a distance of $r_2 = 12 	ext{ mm}$.

Because the line charge is positive, it creates an electric field that points radially outward. The positive charge of the dipole will experience a repulsive force $F_1$ pushing it away from the origin. Conversely, the negative charge will experience an attractive force $F_2$ pulling it towards the origin.

### The Master Equation

To find the net force on the dipole, we first need the formula for the electric field created by an infinite line charge at a distance $r$:

$$E = \frac{2k\lambda}{r}$$

The force on any point charge in an electric field is given by $F = qE$. Therefore, the forces on our two charges are:

$$F_1 = qE_1 = q \left( \frac{2k\lambda}{r_1} ight)$$
$$F_2 = qE_2 = q \left( \frac{2k\lambda}{r_2} ight)$$

Since $r_1 < r_2$, the electric field is stronger at the location of the positive charge. This means the repulsive force $F_1$ is greater than the attractive force $F_2$. The net force $F_{net}$ will be directed away from the origin:

$$F_{net} = F_1 - F_2 = 2k\lambda q \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$$

### Final Calculation

We are given that the net force $F_{net} = 4 	ext{ N}$. Let's substitute all our known values into the equation. Remember to convert the distances from millimeters to meters to maintain SI units!

$$4 = 2 	imes (9 	imes 10^9) 	imes (3 	imes 10^{-6}) 	imes q 	imes \left( \frac{1}{10 	imes 10^{-3}} - \frac{1}{12 	imes 10^{-3}} ight)$$

Let's simplify the constants outside the bracket:

$$4 = 54 	imes 10^3 	imes q 	imes 10^3 \left( \frac{1}{10} - \frac{1}{12} ight)$$

Now, solving the fraction inside the bracket:

$$\frac{1}{10} - \frac{1}{12} = \frac{12 - 10}{120} = \frac{2}{120} = \frac{1}{60}$$

Substituting this back into our equation:

$$4 = 54 	imes 10^6 	imes q 	imes \left( \frac{1}{60} ight)$$

$$4 = 0.9 	imes 10^6 	imes q = 9 	imes 10^5 	imes q$$

Finally, isolating $q$:

$$q = \frac{4}{9 	imes 10^5} = \frac{4}{9} 	imes 10^{-5} 	ext{ C}$$

$$\mathbf{q = 0.444 	imes 10^{-5} 	ext{ C} = 4.44 	imes 10^{-6} 	ext{ C} = 4.44 \mu	ext{C}}$$

The magnitude of the charge on the dipole is exactly **$4.44 \mu	ext{C}$**.

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Visualized Solution

$Visualizing the Setup$

$Electric Field of Line Charge$

$Net Force on Dipole$

$Substituting the Formula$

$Plugging in the Values$

$Simplifying Constants$

$Solving the Fraction$

$Isolating q$

$Final Answer$

$Food for Thought$

The Sigma Insight: Electric Dipole

Analyzing the Setup

The Master Equation

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; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field of Line Charge

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

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

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

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

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

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

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

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