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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

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) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)\)

Equal charges \(q\) are placed at the vertices \(A\) and \(B\) of an equilateral triangle \(\mathrm{ABC}\) of side \(\mathrm{a}\). The magnitude of electric field at the point \(c\) is \(\ldots \ldots \ldots\) (A) \(\left(\mathrm{Kq} / \mathrm{a}^{2}\right)\) (B) \(\left.(\sqrt{3} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (C) \(\left.(\sqrt{2} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (D) $\left[\mathrm{q} /\left(2 \pi \mathrm{t} \varepsilon_{0} \mathrm{a}^{2}\right)\right]$

64 identical drops of mercury are charged simultaneously to the same potential of 10 volt. Assuming the drops to be spherical, if all the charged drops are made to combine to form one large drop, then its potential will be (A) \(100 \mathrm{~V}\) (B) \(320 \mathrm{~V}\) (C) \(640 \mathrm{~V}\) (D) \(160 \mathrm{~V}\)

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be $\ldots . .$ (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

Let $\mathrm{P}(\mathrm{r})\left[\mathrm{Q} /\left(\pi \mathrm{R}^{4}\right)\right] \mathrm{r}$ be the charge density distribution for a solid sphere of radius \(\mathrm{R}\) and total charge \(\mathrm{Q}\). For a point ' \(\mathrm{P}\) ' inside the sphere at distance \(\mathrm{r}_{1}\) from the centre of the sphere the magnitude of electric field is (A) $\left[\mathrm{Q} /\left(4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}\right)\right]$ (B) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(4 \pi \in{ }_{0} \mathrm{R}^{4}\right)\right]$ (C) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(3 \pi \epsilon_{0} \mathrm{R}^{4}\right)\right]$
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