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Chapter 11: Problem 1572

Charges of \(+(10 / 3) \times 10^{-9} \mathrm{C}\) are placed at each of the four corners of a square of side \(8 \mathrm{~cm}\). The potential at the intersection of the diagonals is ...... (A) \(150 \sqrt{2}\) Volt (B) \(900 \sqrt{2}\) Volt (C) \(1500 \sqrt{2}\) Volt (D) \(900 \sqrt{2} \cdot \sqrt{2}\) Volt

Short Answer

Expert verified
The potential at the intersection of the diagonals is \(1800 \sqrt{2}\ \mathrm{V}\).

Step by step solution

01

Calculate distances between the charges and the center of the square.

Since the charges are at the corners of the square, the distance between each charge and the center can be found using the Pythagorean theorem. The length of the diagonal is given by \(d = \sqrt{2} \cdot a\), where \(a\) is the side length. Therefore, the distance between each corner and the center is half of the diagonal length, which can be calculated as follows: \(r = \frac{d}{2} = \frac{\sqrt{2} \cdot a}{2}\).
02

Substitute values for side length and charge.

Now, let's substitute the values for the side length (\(a = 8 \mathrm{cm}\)) and the charge: \(Q = +(10 / 3) \times 10^{-9} \mathrm{C}\). We need to convert the side length to meters, which is \(a = 0.08 \mathrm{m}\).
03

Calculate the distance between each charge and the center.

Now, we can calculate the distance between each charge and the center of the square using the formula from before: \(r = \frac{\sqrt{2} \cdot a}{2} = \frac{\sqrt{2} \cdot 0.08}{2} = 0.0328\sqrt{2}\ \mathrm{m}\).
04

Calculate the potential due to one charge at the center.

Next, we can compute the potential due to a single charge, using the formula \(V = \frac{kQ}{r}\), with \(k = 8.99 \times 10^{9} \frac{\mathrm{N m}^2}{\mathrm{C}^2}\), \(Q = +(10/3) \times 10^{-9} \mathrm{C}\), and \(r = 0.0328\sqrt{2}\ \mathrm{m}\). The potential is given by: \(V_1 = \frac{8.99 \times 10^{9} \frac{\mathrm{N m}^2}{\mathrm{C}^2} \cdot +(10/3) \times 10^{-9} \mathrm{C}}{0.0328\sqrt{2}\ \mathrm{m}} = 450 \sqrt{2} \mathrm{V}\).
05

Sum up the potential from all four charges.

Finally, since the charges are all identical and equidistant from the center of the square, the potential due to each charge contributes the same amount to the total potential. Therefore, we can multiply the potential from one charge by 4 to get the total potential at the center: \(V_{\text{total}} = 4 \times V_1 = 4 \times 450 \sqrt{2} \mathrm{V} = 1800 \sqrt{2}\ \mathrm{V}\). Since none of the given options matches our calculated total potential of \(1800\sqrt{2}\ \mathrm{V}\), we could conclude that there might be an error in the given answer options for this exercise.

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Most popular questions from this chapter

Two point charges repel each other with a force of \(100 \mathrm{~N}\). One of the charges is increased by \(10 \%\) and other is reduced by \(10 \%\). The new force of repulsion at the same distance would be \(\ldots \ldots \mathrm{N}\). (A) 121 (B) 100 (C) 99 (D) 89

The plates of a parallel capacitor are charged up to \(100 \mathrm{~V}\). If $2 \mathrm{~mm}$ thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by \(1.6 \mathrm{~mm}\) the dielectric constant of the plate is (A) 5 (B) 4 (C) \(1.25\) (D) \(2.5\)

A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(\ell\). The maximum possible flux of the electric field through the cube will be ....... (A) \(\sqrt{3}\left(\lambda \ell / \in_{0}\right)\) (B) \(\left(\lambda \ell / \in_{0}\right)\) (C) \(\sqrt{2}\left(\lambda \ell / \in_{0}\right)\) (D) \(\left[\left(6 \lambda \ell^{2}\right) / \epsilon_{0}\right]\)

A \(5 \mu \mathrm{F}\) capacitor is charged by a \(220 \mathrm{v}\) supply. It is then disconnected from the supply and is connected to another uncharged $2.5 \mu \mathrm{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? (A) \(0.02 \mathrm{~J}\) (B) \(0.121 \mathrm{~J}\) (C) \(0.04 \mathrm{~J}\) (D) \(0.081 \mathrm{~J}\)

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) \(-(9 / 4) \mathrm{J}\)
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