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Chapter 17: Problem 80

Two point charges \(-q\) and \(+q / 2\) are situated at the origin and at the point \((a, 0,0)\), respectively. The point along the X-axis, whereas the electric field vanished, is (a) \(x=\frac{\sqrt{2} a}{\sqrt{2}-1}\) (b) \(X=\sqrt{2} a-\sqrt{2}-1\) (c) \(x=(\sqrt{2}-1) \sqrt{2} a\) (d) None of these

Short Answer

Expert verified
The correct choice is (c) \(x=(\sqrt{2}-1) \sqrt{2} a\).

Step by step solution

01

Understanding the electric field contributions

Consider two charges: \(-q\) at the origin \((0,0,0)\) and \(+\frac{q}{2}\) at \((a,0,0)\). The electric field \(E\) due to a point charge \(q\) at a distance \(r\) can be calculated using the formula \(E = \frac{kq}{r^2}\), where \(k\) is the Coulomb's constant.
02

Set up the equation for the electric field

We want to find the point along the X-axis where the total electric field is zero. Let this point be \(x\). The electric field at \(x\) from the charge \(-q\) is \(E_1 = \frac{-kq}{x^2}\). The electric field at \(x\) from the charge \(+\frac{q}{2}\) is \(E_2 = \frac{k(q/2)}{(x-a)^2}\). Hence, the total electric field \(E_{total} = E_1 + E_2\). Set \(E_{total} = 0\) for there to be a point where the electric field vanishes.
03

Solve the equation for zero total electric field

Combine the expressions for \(E_1\) and \(E_2\):\[\frac{-kq}{x^2} + \frac{k(q/2)}{(x-a)^2} = 0\]This simplifies to:\[-\frac{q}{x^2} + \frac{q}{2(x-a)^2} = 0\]Multiply both sides by \(x^2(x-a)^2\) to clear denominators:\[-(x-a)^2 + \frac{x^2}{2} = 0\]
04

Further solve the equation

Expand and simplify the equation from Step 3:\[-x^2 + 2ax - a^2 + \frac{x^2}{2} = 0\]Combine like terms:\[-\frac{x^2}{2} + 2ax - a^2 = 0\]Multiply the entire equation by -2 to simplify:\[x^2 - 4ax + 2a^2 = 0\]
05

Use the quadratic formula

Solve the quadratic equation \(x^2 - 4ax + 2a^2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):Here, \(a = 1\), \(b = -4a\), \(c = 2a^2\).\[x = \frac{4a \pm \sqrt{(-4a)^2 - 4 \times 1 \times 2a^2}}{2}\]Simplify:\[x = \frac{4a \pm \sqrt{16a^2 - 8a^2}}{2}\]\[x = \frac{4a \pm \sqrt{8a^2}}{2}\]\[x = \frac{4a \pm 2\sqrt{2}a}{2}\]\[x = 2a \pm \sqrt{2}a\]
06

Identify the acceptable root

The two potential solutions are \(x = 2a + \sqrt{2}a\) and \(x = 2a - \sqrt{2}a\). Since we are looking for the point along the X-axis where the electric field vanishes and must lie between 0 and \(a\), eliminate \(2a + \sqrt{2}a\) as it is greater than \(a\). Thus, the correct distance where the electric field vanishes is:\[x = (2 - \sqrt{2})a\].

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Key Concepts

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

Point Charge
A point charge is a hypothetical charge that has all its charge concentrated at one point in space. It serves as a simplified model to understand the electric field effects in physics. Imagine a tiny, invisible dot where the entire charge resides. This concept helps us analyze how charges interact with each other and with electric fields in space.

In real-world scenarios, point charges are valuable because they allow us to apply mathematical models to predict the behavior of electric fields. Even though no charge is infinitely small, the concept of a point charge is useful for calculations when the size of the charge is much smaller than the distances involved in the analysis.
  • Point charges are used extensively in physics to model atoms and subatomic particles.
  • They help simplify complex systems into manageable mathematical problems.
Electric Field Zero
The electric field zero point is the location where the total electric field due to multiple charges becomes zero. This is typically a point along the path connecting two or more charges where their electric fields cancel each other out.

Finding this point involves calculating the electric fields generated by individual charges and summing them to zero. A thorough understanding of the conditions under which electric fields cancel is crucial.
  • You need to consider the direction and magnitude of each contributing field.
  • This concept is especially important in understanding superposition and balance in fields.
Determining where the electric field is zero can have practical applications, such as designing circuits or other electrical engineering tasks where control of electric fields is crucial.
Coulomb's Law
Coulomb's Law is a fundamental principle that describes how two point charges interact with each other. It states that the electric force between two point charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them. The direction of the force is along the line joining the charges.

Mathematically, Coulomb's Law is expressed as:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:
  • \(F\) is the magnitude of the force between the two charges,
  • \(q_1\) and \(q_2\) are the amounts of the two charges,
  • \(r\) is the distance between the centers of the two charges,
  • \(k\) is Coulomb's constant \((8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)\).
Coulomb's Law not only predicts the magnitude and direction of the force but also provides insights into how charged objects influence each other within an electric field.
Quadratic Equation in Physics
Quadratic equations appear frequently in physics problems, where the relationship between variables can result in a second-degree polynomial equation. Solving these equations allows us to find critical points in physical systems, such as maximums or minimums, points of equilibrium, or in this case, locations where an electric field vanishes.

The standard form is expressed as:\[ ax^2 + bx + c = 0 \]The solutions for the quadratic equation can be found using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • In the context of electric fields, it helps compute positions where forces or fields balance each other.
  • Understanding these equations is crucial in analyzing the dynamics of systems influenced by quadratic relationships.
This mathematical tool is indispensable for physicists when exploring more complicated systems, like those involving multiple point charges.

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Most popular questions from this chapter

In a region of space having a unifrom electric field \(E\), a hemispherical bowl of radius \(r\) is placed. The electric flux \phi through the bowl is (a) \(2 \pi r E\) (b) \(4 \pi r^{2} E\) (c) \(2 \pi r^{2} E\) (d) \(\pi r^{2} E\)

Two particles \(A\) and \(B\) having charges \(8 \times 10^{-6} \mathrm{C}\) and \(-2 \times 10^{-6} \mathrm{C}\) respectively, are held fixed with a separation \(20 \mathrm{~cm} .\) Where should a third charged particle be placed so that it does not experience a net electric force? (a) \(0.2 \mathrm{~m}\) (b) \(0.5 \mathrm{~m}\) (c) \(0.6 \mathrm{~m}\) (d) \(0.1 \mathrm{~m}\)

Two point charges \(+q\) and \(-q\) are held fixed at \((-d, 0)\) and \((d, 0)\) respectively of a \((x, y)\) coordinate system, then (a) the electric field \(\mathrm{E}\) at all points on the \(x\)-axis has the same direction (b) \(\mathrm{E}\) at all points on the \(y\)-axis is along \(\hat{i}\) (c) work has to be done in bringing a test charge from infinity to the origin (d) the dipole moment is 2 qd directed along \(\hat{i}\)

If the electric flux entering and leaving an enclosed surface are \(\phi_{1}\) and \(\phi_{2}\) respectively, then charge enclosed in closed surface is (a) \(\frac{\phi_{2}-\phi_{1}}{\varepsilon_{0}}\) (b) \(\frac{\phi_{1}+\phi_{2}}{\varepsilon_{0}}\) (c) \(\frac{\phi_{1}-\phi_{2}}{\varepsilon_{0}}\) (d) \(\varepsilon_{0}\left(\phi_{2}-\phi_{1}\right)\)

A long charged cylinder of linear charged density \(\lambda\) is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders? [NCERT] (a) \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\) (b) \(\frac{\lambda r}{\sqrt{2} \pi \varepsilon_{0}}\) (c) \(\frac{\lambda}{\sqrt{2} \pi \varepsilon_{0} r}\) (d) None of these
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