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Chapter 11: Problem 1592

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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Most popular questions from this chapter

A particle having a charge of \(1.6 \times 10^{-19} \mathrm{C}\) enters between the plates of a parallel plate capacitor. The initial velocity of the particle is parallel to the plates. A potential difference of \(300 \mathrm{v}\) is applied to the capacitor plates. If the length of the capacitor plates is $10 \mathrm{~cm}\( and they are separated by \)2 \mathrm{~cm}$, Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is \(12 \times 10^{-24} \mathrm{~kg}\). (A) \(10^{4}(\mathrm{~m} / \mathrm{s})\) (B) \(10^{2}(\mathrm{~m} / \mathrm{s})\) (C) \(10^{-1}(\mathrm{~m} / \mathrm{s})\) (D) \(10^{3}(\mathrm{~m} / \mathrm{s})\)

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases \(4.5\) times in comparison with the initial value. The ratio of the initial charges of the balls is ....... (A) \(4: 1\) (B) \(6: 1\) (C) \(3: 1\) (D) \(2: 1\)

The electric Potential \(\mathrm{V}\) at any Point $0(\mathrm{x}, \mathrm{y}, \mathrm{z}\( all in meters \))\( in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}\( volt. The electric field at the point \)(1 \mathrm{~m}, 0.2 \mathrm{~m})\( in volt meter is \)\ldots \ldots .$ (A) 8 , along negative \(\mathrm{x}\) - axis (B) 8 , along positives \(\mathrm{x}\) - axis (C) 16 , along negative \(\mathrm{x}\) -axis (D) 16 , along positives \(\mathrm{x}\) -axis

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and work done by the battery will be (A) \((1 / 2)\) (B) \((2 / 1)\) (C) 1 (D) \((1 / 4)\)

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) $\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]$ (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) $\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]$ (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)
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