/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 If two electric charges \(q\) an... [FREE SOLUTION] | 91Ó°ÊÓ

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If two electric charges \(q\) and \(-2 q\) are placed at a distance 6a apart, then the locus of point in the plane of charge, where the field potential is zero, is : (a) \(x^{2}+2 y^{2}-4 a x-12 a^{2}=0\) (b) \(2 x^{2}+y^{2}+4 a x-12 a^{2}=0\) (c) \(x^{2}+y^{2}+4 a x-12 a^{2}=0\) (d) \(x^{2}+y^{2}+8 a x+12 a^{2}=0\)

Short Answer

Expert verified
The locus where the field potential is zero is given by the equation (c) \(x^2 + y^2 + 4ax - 12a^2=0\).

Step by step solution

01

Understand the Problem

We are given two charges, \(q\) and \(-2q\), placed 6a units apart. We need to find the locus where the electric potential is zero. The electric potential \(V\) at a point due to a charge \(Q\) at distance \(r\) is given by \(V = \frac{kQ}{r}\), where \(k\) is a constant.
02

Determine the Position of Charges

Let the positive charge \(q\) be placed at the origin \((0,0)\) and the negative charge \(-2q\) be placed at \((6a, 0)\) on the x-axis. The point where we consider the potential, \((x,y)\), will have distances \(r_1\) and \(r_2\) from \(q\) and \(-2q\), respectively.
03

Write the Expression for Electric Potential

The net potential \(V\) at the point \((x,y)\) is given by:\[ V = \frac{kq}{r_1} - \frac{2kq}{r_2} \]where \(r_1 = \sqrt{x^2 + y^2}\) and \(r_2 = \sqrt{(x - 6a)^2 + y^2}\). For the potential to be zero, we need:\[ \frac{kq}{r_1} = \frac{2kq}{r_2} \]
04

Solve for Zero Potential Condition

Cancel out \(kq\) from both sides:\[ \frac{1}{r_1} = \frac{2}{r_2} \]This leads to:\[ r_2 = 2r_1 \]
05

Substitute Distances and Simplify

Substitute \(r_1 = \sqrt{x^2 + y^2}\) and \(r_2 = \sqrt{(x - 6a)^2 + y^2}\):\[ \sqrt{(x - 6a)^2 + y^2} = 2 \sqrt{x^2 + y^2} \]Square both sides to eliminate the square roots:\[ (x - 6a)^2 + y^2 = 4(x^2 + y^2) \]
06

Simplify the Equation

Expand the expression on the left:\[ x^2 - 12ax + 36a^2 + y^2 = 4x^2 + 4y^2 \]Rearrange and combine like terms:\[ 3x^2 + 3y^2 + 12ax - 36a^2 = 0 \]Divide the entire equation by 3 to simplify:\[ x^2 + y^2 + 4ax - 12a^2 = 0 \]
07

Check the Available Options

Compare the derived equation \(x^2 + y^2 + 4ax - 12a^2 = 0\) with the options provided and identify the correct match. This matches with option (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Charges
Electric charges are fundamental properties of matter, and they come in two types: positive and negative. Like charges repel each other, while opposite charges attract. The charge of an electron is negative and is often noted as \(-e\), while a proton carries a positive charge of \(+e\). Several basic principles govern electric charges:
  • Quantization: Charge is quantized, meaning it exists in discrete amounts. The smallest unit of charge is that of an electron or proton, referred to as elementary charge.
  • Conservation: In a closed system, the total electric charge remains constant over time. This means the algebraic sum of all charges remains unchanged.
In the exercise given, the two charges involved are \(q\) and \(-2q\). These charges are situated a fixed distance apart, and the principles of electric potential apply to them extensively.
Locus of Zero Potential
The locus of zero potential refers to the set of all points in a field where the electric potential is zero. When dealing with multiple charges, each charge contributes to the potential at a point. The condition for zero potential depends on the spatial arrangement and magnitudes of these charges.In our exercise, we have a charge \(q\) at the origin and a charge \(-2q\) at some distance away. The locus of points with zero potential will depend on their distances from these charges. By equating the potentials from the charges at a point to zero, we find the equation of the locus which satisfies the zero potential condition. Through algebraic manipulation, this locus is mathematically expressed, leading us to the correct option (c) of the exercise.
Distance and Potential Relationship
The electric potential \(V\) at a point due to a point charge \(Q\) is given by \(V = \frac{kQ}{r}\), where \(k\) is Coulomb's constant and \(r\) is the distance from the charge to the point. This relationship highlights the inverse nature of potential and distance: as the distance increases, the potential decreases.For our two charges, this relationship forms the basis for establishing the zero potential condition: \(q\)'s potential is balanced by \(-2q\)'s potential at some point so that the net potential is zero. Using geometric distance formulas, we identify how the potentials interplay due to their differing distances. Solving through the condition \(\frac{kq}{r_1} = \frac{2kq}{r_2}\) leads to this locus equation. Understanding the inverse distance-potential relationship is key to interpreting the locus of zero potential.
Quadratic Equations in Physics
Quadratic equations frequently appear in physics, often in contexts that involve geometric or dynamic systems. They are expressed in the form \(ax^2 + bx + c = 0\) and can describe paths, forces, or potentials in systems.In this case, simplifying the derived locations of zero potential using coordinate systems translates into a quadratic form: \(x^2 + y^2 + 4ax - 12a^2 = 0\). This equation describes a conic section, indicating the path or shape that would satisfy the zero potential condition between the charges. Quadratic equations thus capture complex relationships compactly, linking geometrical loci to physical scenarios.

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Most popular questions from this chapter

A \(\mathrm{Na}^{+}\) ion is moving through an evacuated vessel in the positive \(x\) -direction with speed of \(10^{7} \mathrm{~m} / \mathrm{s}\). At \(x=0, y=0\) it enters an electric field of \(500 \mathrm{~V} / \mathrm{cm}\) in the positive \(x\) -direction. Its position \((x, y)\) after \(10^{-6} \mathrm{~s}\) is: (a) \((10,5)\) (b) \((1,0.4)\) (c) \((10,0.1)\) (d) \((10,8)\)

Mark correct statement(s): (a) When two charged bodies attract each other, then two bodies must have opposite nature of charges (b) When two charged bodies attract each other, then they may have the same nature of charge (c) Potential difference between two points lying in a uniform electric field may be equal to zero (d) Both (b) and (c) are correct

If \(Q\) charge is given to a spherical sheet of radius \(R\), the energy of the system is: (a) \(\frac{Q^{2}}{8 \pi \varepsilon_{0} R}\) (b) \(\frac{Q^{2}}{4 \pi \varepsilon_{0} R}\) (c) \(\frac{Q^{2}}{15 \pi \varepsilon_{0} R}\) (d) none of these

A particle of mass \(2 \mathrm{~g}\) and charge \(1 \mu \mathrm{C}\) is held at rest on a frictionless horizuntal surface at a distance of \(1 \mathrm{~m}\) from a fixed charge \(1 \mathrm{mC}\). If the particle is released, it will be repelled. The speed of the particle when it is at a distance of \(10 \mathrm{~m} \leqslant \mathrm{~m}\) the fixed charge is: (a) \(100 \mathrm{~m} / \mathrm{s}\) (b) \(90 \mathrm{~m} / \mathrm{s}\) (c) \(60 \mathrm{~m} / \mathrm{s}\) (d) \(45 \mathrm{~m} / \mathrm{s}\)

A charge \(+Q\) is uniformly distributed over a thin ring of the radius \(R\). The velocity of an electron at the moment when it passes through the centre \(O\) of the ring, if the electron was initially at rest at a point \(A\) which is very far always from the centre and on the axis of the ring is : (a) \(\sqrt{\left(\frac{2 K Q e}{m R}\right)}\) (b) \(\sqrt{\left(\frac{K Q e}{m}\right)}\) (c) \(\sqrt{\left(\frac{K m e}{Q R}\right)}\) (d) \(\sqrt{\left(\frac{K Q e}{m R}\right)}\)

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