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4 Points charges each \(+q\) is placed on the circumference of a circle of diameter \(2 \mathrm{~d}\) in such a way that they form a square. The potential at the centre is \(\ldots \ldots .\) (A) 0 (B) \((4 \mathrm{kd} / \mathrm{q})\) (C) \((\mathrm{kd} / 4 \mathrm{q})\) (D) \((4 \mathrm{kq} / \mathrm{d})\)

Short Answer

Expert verified
The electric potential at the center of the circle is \(\frac{4kq}{d}\), which corresponds to option (D).

Step by step solution

01

Calculate side length of the square

Since the square is inscribed in a circle with diameter \(2d\), the diagonal of the square = diameter of the circle. We'll use the Pythagorean theorem to find the side length of the square: \(s^2 + s^2 = (2d)^2\) Solving for \(s\), we get: \(s = d\sqrt{2}\)
02

Calculate distance of each charge to the center

We'll use the Pythagorean theorem again to find the distance from each corner of the square (where the charges are placed) to the center of the circle. As the charges are equidistant from the center, we can find the distance of any one charge to the center. \(\frac{s}{2}^2 + \frac{s}{2}^2 = r^2\) Substituting the value of \(s\), we get: \(\frac{d\sqrt{2}}{2}^2 + \frac{d\sqrt{2}}{2}^2 = r^2\) On solving for \(r\), we get \(r = d\) #Step 2: Calculate the electric potential at the center of the circle#
03

Calculate the potential due to each charge

Using the formula for electric potential due to a point charge (\(V = k\frac{q}{r}\)), we can calculate the potential at the center due to one point charge: \(V_{1charge} = k\frac{q}{d}\)
04

Calculate the total potential at the center

Since there are 4 point charges and electric potential is a scalar quantity, we simply sum the contributions from each charge: \(V_{total} = 4 \times V_{1charge} = 4 \times k\frac{q}{d} = \frac{4kq}{d}\) So, the electric potential at the center of the circle is \(\frac{4kq}{d}\), which matches option (D).

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Key Concepts

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

Point Charges
Point charges are a foundational concept in physics, particularly in electrostatics. They refer to objects or particles that have an electric charge, such as electrons or protons. These charges cause an electric field around them, which influences other charges in the vicinity. The magnitude of this influence depends on both the size of the charge and the distance from the charge.
  • For a positive point charge, the electric field radiates outward.
  • For a negative point charge, the electric field points inward.
In the given exercise, four point charges of charge "+q" each are placed at the corners of a square, which is inscribed in a circle. These charges create an electric field that influences the potential at the center of the circle, allowing us to calculate the total electric potential from all four sources.
Pythagorean Theorem
The Pythagorean Theorem is a mathematical principle used to relate the sides of a right triangle. It's particularly handy when we need to determine distances, like in the exercise where we're dealing with a square within a circle. It states that in a right triangle, the sum of the squares of the two shorter sides equals the square of the hypotenuse.
  • The formula is: \(a^2 + b^2 = c^2\).
In this problem, by treating the square's diagonal as a hypotenuse, and its sides as the shorter sides, we apply the theorem to find the lengths we need to solve for potentials.
Scalar Quantity
A scalar quantity is a physical quantity that is described by a single number, or magnitude, without any direction. It contrasts with vector quantities, which also have a direction. Examples of scalars include temperature, mass, and in this case, electric potential.
Electric potential due to point charges is additive and directionless, making it a scalar quantity. This characteristic simplifies calculations because the total electric potential at any point can be found simply by summing individual potentials, as was done in the exercise. In the final step of the solution, the scalar nature enabled straightforward addition of the potential from each of the four point charges.

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Most popular questions from this chapter

Two charged spheres of radii \(R_{1}\) and \(R_{2}\) having equal surface charge density. The ratio of their potential is ..... (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)\) (B) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)\)

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q\). The potential difference between the centers of the two rings is \(\ldots .\) (A) 0 (B) \(\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]\) (C) \(\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.\) (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)

A charged particle of mass \(1 \mathrm{~kg}\) and charge \(2 \mu \mathrm{c}\) is thrown from a horizontal ground at an angle \(\theta=45^{\circ}\) with speed \(20 \mathrm{~m} / \mathrm{s}\). In space a horizontal electric field \(\mathrm{E}=2 \times 10^{7} \mathrm{~V} / \mathrm{m}\) exist. The range on horizontal ground of the projectile thrown is \(\ldots \ldots \ldots\) (A) \(100 \mathrm{~m}\) (B) \(50 \mathrm{~m}\) (C) \(200 \mathrm{~m}\) (D) \(0 \mathrm{~m}\)

Two identical metal plates are given positive charges \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{2}\left(<\mathrm{Q}_{1}\right)\) respectively. If they are now brought close to gather to form a parallel plate capacitor with capacitance \(\mathrm{c}\), the potential difference between them is (A) \(\left[\left(Q_{1}+Q_{2}\right) /(2 c)\right]\) (B) \(\left[\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right) / \mathrm{c}\right]\) (C) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) /(2 \mathrm{c})\right]\) (D) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) / \mathrm{c}\right]\)

Two identical charged spheres suspended from a common point by two massless strings of length \(\ell\) are initially a distance d \((d<<\ell)\) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the spheres approach each other with a velocity \(\mathrm{v}\). Then function of distance \(\mathrm{x}\) between them becomes \(\ldots \ldots\) (A) \(v \propto x\) (B) \(\mathrm{v} \propto \mathrm{x}^{(-1 / 2)}\) (C) \(\mathrm{v} \propto \mathrm{x}^{-1}\) (D) \(\mathrm{v} \propto \mathrm{x}^{(1 / 2)}\)

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