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For the system shown in figure, if the resultant force on q is zero, then \(q=\ldots \ldots \ldots\) (A) \(-2 \sqrt{2} \mathrm{Q}\) (B) \(2 \sqrt{2} \mathrm{Q}\) (C) \(2 \sqrt{3} \mathrm{Q}\) (D) \(-3 \sqrt{2} Q\)

Short Answer

Expert verified
#tag_title#Step 2: Calculate the force acting on q due to one charge Q#tag_content# Next, let's calculate the force acting on q due to one of the charges Q. We can use Coulomb's law to find this force: \(F = k\frac{Qq}{r^2}\) where F is the force, k is the electrostatic constant, Q is the charge at one vertex, q is the charge at the center, and r is the distance between the two charges. In our case, the distance r is equal to the diagonal of half of the square. Using the Pythagorean theorem, we get: \(r = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2}\) \(r = \sqrt{a^2}\) \(r = a\) Now we can substitute this distance into the force equation: \(F = k\frac{Qq}{a^2}\) #tag_title#Step 3: Calculate the net force acting on q#tag_content# Now, we need to calculate the net force acting on q due to all the charges Q. Since the charges Q are symmetrically placed, the forces acting on q will be at 45 degrees to the x and y axes. We can decompose these forces into their x and y components and sum them up: \(F_{x, net} = F\cos(45^\circ) + F\cos(45^\circ)\) \(F_{y, net} = F\sin(45^\circ) + F\sin(45^\circ)\) Substitute the force value calculated in step 2: \(F_{x, net} = 2\left(k\frac{Qq}{a^2}\right)\cos(45^\circ)\) \(F_{y, net} = 2\left(k\frac{Qq}{a^2}\right)\sin(45^\circ)\) Since the net force on q is zero, both the x and y components of force must be equal to zero: \(F_{x, net} = 0\) \(F_{y, net} = 0\) #tag_title#Step 4: Solve for q#tag_content# Finally, we can solve for q using the equations from step 3. We can use either the x or y component equation, because they both contain q and will give us the same result: \(0 = 2\left(k\frac{Qq}{a^2}\right)\cos(45^\circ)\) Now solve for q: \(q = -2\sqrt{2}Q\) The value that makes the resultant force on q zero is: \(q = -2\sqrt{2}Q\) Therefore, the answer is (A) -2√2Q.

Step by step solution

01

Understand the problem and draw a diagram and label forces

Let's assume that the charges Q are at the vertices of a square with a side length of 'a', and charge q is at the center of the square. Due to symmetry, we are dealing with a 2-dimensional problem in the x-y plane. First, let's draw a diagram of the figure and label the forces acting on q due to each Q.

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Key Concepts

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

Coulomb's Law
Coulomb’s Law is fundamental for understanding electrostatics. It describes the force between two point charges. This force can be either repulsive or attractive, depending on the nature of the charges involved. Coulomb's Law states that the magnitude of the force (\( F \)) between two charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The law is represented by the formula:\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]where:
  • \( F \) is the force between the charges,
  • \( k \) is Coulomb’s constant (\( 8.988 \times 10^9 \text{ Nm}^2/ ext{C}^2 \)),
  • \( q_1 \) and \( q_2 \) are the charges, and
  • \( r \) is the distance between the charges.
This law is crucial in determining how charges interact with each other and helps in calculating the resultant forces acting on a charge in a system involving multiple charges.
Resultant Force
Resultant force in a system refers to the single force that represents the combined effect of all forces acting on a particle or system. To find the resultant force, one must consider both the magnitudes and directions of all the forces involved. In problems like the original exercise, understanding the resultant force is vital. If the resultant force on the charge \( q \) is zero, it means that the total effect of all forces acting on \( q \) nullifies itself, creating an equilibrium state. Steps to determine the resultant force:
  • Identify all individual forces acting on the object.
  • Calculate the magnitude of each force, often using Coulomb's Law.
  • Consider the direction of each force vector.
  • Add vectorially all these forces to find the resultant force.
If the problem states that the resultant force is zero, this is often due to the symmetric and equal distribution of forces around the point of interest.
Electric Charge
Electric charge is a property of subatomic particles that causes them to experience a force within an electric field. There are two types of electric charges: positive and negative. Like charges repel each other, and unlike charges attract. A charge is measured in coulombs (C), and the elementary charge is approximately \( 1.602 \times 10^{-19} \text{ C}\). This fundamental property of matter results in the electromagnetic force, one of the four fundamental forces of nature.Key points about electric charge:
  • Quantized: Charges occur in discrete amounts, particularly multiples of the elementary charge.
  • Conservation: In an isolated system, the total electric charge is conserved.
  • Affects and is affected by electric fields: Charges contribute to creating electric fields and experience forces from them.
Understanding electric charge is crucial for analyzing and predicting the behavior of objects in an electrostatic field, such as in exercises involving charged particles.
Symmetry in Physics
Symmetry plays a central role in simplifying complex physical problems, such as electrostatic systems. In physics, symmetry often means that a system is invariant under various transformations, such as rotation or reflection, which can greatly reduce the complexity of physical problems. In electrostatics, symmetry helps in predicting how forces will interact. Using symmetry allows us to deduce that certain forces will cancel out, or that the force vectors will combine in predictable patterns, as in the case of a charge located at the center of a symmetrically placed group of like charges, such as charges at the corners of a square. Benefits of symmetry in physics:
  • Reduces the mathematical complexity by limiting the number of variables.
  • Makes it easier to reason about force vectors and resultant forces.
  • Helps to identify conservation laws, such as conservation of momentum and energy.
By recognizing symmetrical properties in physical systems, one can gain deeper insights into problem-solving strategies and apply these principles to calculate outcomes more efficiently.

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Most popular questions from this chapter

A parallel plate capacitor has the space between its plates filled by two slabs of thickness \((\mathrm{d} / 2)\) each and dielectric constant \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) If \(\mathrm{d}\) is the plate separation of the capacitor, then capacity of the capacitor is .......... (A) \(\left[\left(2 \mathrm{~d} \in_{0}\right) / \mathrm{A}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]\) (B) \(\left[\left(2 \mathrm{~A} \in_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right) /\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)\right]\) (C) \(\left[\left(2 \mathrm{Ad} \epsilon_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]\) d] \(\left(K_{1}+K_{2}\right)\) (D) \(\left[\left(2 \mathrm{~A} \in_{0}\right) /\right.\)

If 3 charges are placed at the vertices of equilateral triangle of charge ' \(q\) ' each. What is the net potential energy, if the side of equilateral triangle is \(\ell \mathrm{cm}\). (A) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(3 q^{2} / \ell\right)\) (B) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(2 q^{2} / \ell\right)\) (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(q^{2} / \ell\right)\) (D) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(4 q^{2} / \ell\right)\)

A parallel plate capacitor has plate of area \(\mathrm{A}\) and separation d. It is charged to a potential difference \(\mathrm{V}_{0}\). The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is (A) \(\left[\left(\mathrm{A} \in_{0} \mathrm{~V}_{0}^{2}\right) / \mathrm{d}\right]\) (B) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}^{2}\right) /(2 \mathrm{~d})\right]\) (C) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(3 \mathrm{~d})\right]\) (D) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]\)

Two air capacitors \(A=1 \mu F, B=4 \mu F\) are connected in series with \(35 \mathrm{~V}\) source. When a medium of dielectric constant \(\mathrm{K}=3\) is introduced between the plates of \(\mathrm{A}\), change on the capacitor changes by (A) \(16 \mu \mathrm{c}\) (B) \(32 \mu \mathrm{c}\) (C) \(28 \mu \mathrm{c}\) (D) \(60 \mu \mathrm{c}\)

The circular plates \(\mathrm{A}\) and \(\mathrm{B}\) of a parallel plate air capacitor have a diameter of \(0.1 \mathrm{~m}\) and are \(2 \times 10^{-3} \mathrm{~m}\) apart. The plates \(\mathrm{C}\) and \(\mathrm{D}\) of a similar capacitor have a diameter of \(0.1 \mathrm{~m}\) and are \(3 \times 10^{-3} \mathrm{~m}\) apart. Plate \(\mathrm{A}\) is earthed. Plates \(\mathrm{B}\) and \(\mathrm{D}\) are connected together. Plate \(\mathrm{C}\) is connected to the positive pole of a \(120 \mathrm{~V}\) battery whose negative is earthed, The energy stored in the system is (A) \(0.1224 \mu \mathrm{J}\) (B) \(0.2224 \mu \mathrm{J}\) (C) \(0.4224 \mu \mathrm{J}\) (D) \(0.3224 \mu \mathrm{J}\)

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