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Chapter 11: Problem 1583

A sphere of radius \(1 \mathrm{~cm}\) has potential of \(8000 \mathrm{v}\), then energy density near its surface will be ...... (A) \(64 \times 10^{5}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (B) \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (C) \(8 \times 10^{3}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (D) \(32\left(\mathrm{~J} / \mathrm{m}^{3}\right)\)

Short Answer

Expert verified
The short answer is: \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\).

Step by step solution

01

Calculate the electric field near the surface

To find the electric field (E) near the surface of the sphere, we can use the formula: E = \( \frac{V}{r} \), where V is the potential, and r is the radius. Given, V = 8000 V, and r = \(1 \times 10^{-2}\) m (1 cm is converted to meters) Now, let's calculate E: E = \( \frac{8000}{1 \times 10^{-2}} \) = \(8 \times 10^{5} \frac{V}{m} \)
02

Calculate the energy density near the surface

Now that we have the value for the electric field (E), we can find the energy density (u) using the formula: u = \( \frac{1}{2}ε_{0}E^{2} \), where \(ε_{0}\) is the vacuum permittivity (\( 8.854 \times 10^{-12} \frac{C^2}{N.m^2} \)) Let's calculate u: u = \( \frac{1}{2}(8.854 \times 10^{-12})(8 \times 10^{5})^{2} \) = \(2.83 \times 10^{6} \frac{J}{m^3} \) After calculating the energy density, we can see that the correct option is: (B) \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\)

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Most popular questions from this chapter

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) \(0.3224 \mu \mathrm{J}\)

There are 10 condensers each of capacity \(5 \mu \mathrm{F}\). The ratio between maximum and minimum capacities obtained from these condensers will be (A) \(40: 1\) (B) \(25: 5\) (C) \(60: 3\) (D) \(100: 1\)

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) \((\mathrm{Q} / \sqrt{3})\)

Point charges \(4 \mu \mathrm{c}\) and \(2 \mu \mathrm{c}\) are placed at the vertices \(\mathrm{P}\) and Q of a right angle triangle \(P Q R\) respectively. \(Q\) is the right angle, \(\mathrm{PR}=2 \times 10^{-2} \mathrm{~m}\) and \(\mathrm{QR}=10^{-2} \mathrm{~m}\). The magnitude and direction of the resultant electric field at \(\mathrm{R}\) is \(\ldots \ldots\) (A) \(4.28 \times 10^{9} \mathrm{NC}^{-1}, 45^{\circ}\) (B) \(2.38 \times 10^{8} \mathrm{NC}^{-1}, 40.9^{\circ}\) (C) \(1.73 \times 10^{4} \mathrm{NC}^{-1}, 34.7^{\circ}\) (D) \(4.9 \times 10^{10} \mathrm{NC}^{-1}, 34.7^{\circ}\)

An electric dipole is placed along the \(\mathrm{x}\) -axis at the origin o. \(\mathrm{A}\) point \(P\) is at a distance of \(20 \mathrm{~cm}\) from this origin such that OP makes an angle \((\pi / 3)\) with the x-axis. If the electric field at P makes an angle \(\theta\) with the x-axis, the value of \(\theta\) would be \(\ldots \ldots \ldots\) (A) \((\pi / 3)+\tan ^{-1}(\sqrt{3} / 2)\) (B) \((\pi / 3)\) (C) \((2 \pi / 3)\) (D) \(\tan ^{-1}(\sqrt{3} / 2)\)
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