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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

Two Points \(P\) and \(Q\) are maintained at the Potentials of \(10 \mathrm{v}\) and \(-4 \mathrm{v}\), respectively. The work done in moving 100 electrons from \(\mathrm{P}\) to \(\mathrm{Q}\) is \(\ldots \ldots \ldots\) (A) \(2.24 \times 10^{-16} \mathrm{~J}\) (B) \(-9.60 \times 10^{-17} \mathrm{~J}\) (C) \(-2.24 \times 10^{-16} \mathrm{~J}\) (D) \(9.60 \times 10^{-17} \mathrm{~J}\)

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) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(4 q^{2} / \ell\right)$

Two point charges \(100 \mu \mathrm{c}\) and \(5 \mu \mathrm{c}\) are placed at points \(\mathrm{A}\) and \(B\) respectively with \(A B=40 \mathrm{~cm}\). The work done by external force in displacing the charge \(5 \mu \mathrm{c}\) from \(\mathrm{B}\) to \(\mathrm{C}\) where \(\mathrm{BC}=30 \mathrm{~cm}\), angle \(\mathrm{ABC}=(\pi / 2)\) and \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\) \(=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{c}^{2}\). (A) \(9 \mathrm{~J}\) (B) \((9 / 25) \mathrm{J}\) (C) \((81 / 20) \mathrm{J}\) (D) \(-(9 / 4) \mathrm{J}\)

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) \((1 / 4)\)

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)
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