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Chapter 11: Problem 1597

A small conducting sphere of radius \(r\) is lying concentrically inside a bigger hollow conducting sphere of radius \(R\). The bigger and smaller sphere are charged with \(\mathrm{Q}\) and \(\mathrm{q}(\mathrm{Q}>\mathrm{q})\) and are insulated from each other. The potential difference between the spheres will be \(\ldots \ldots\) (A) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(Q / R)]\) (B) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(q / R)]\) (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(Q / R)+(q / r)]\) (D) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(\mathrm{q} / \mathrm{R})-(\mathrm{Q} / \mathrm{r})]$

Short Answer

Expert verified
The potential difference between the spheres is: \[V = \left[\frac{1}{4 \pi \epsilon_0}\right] \left[\frac{Q}{R}-\frac{q}{r}\right]\]

Step by step solution

01

Find potential at the surface of the inner sphere

To find the potential at the surface of the inner sphere, we use the formula V = kQ/r, where Q = q (charge on the inner sphere) and r is its radius. V_inner = k * (q / r)
02

Find potential at the surface of the outer sphere

Similarly, we will now find the potential at the surface of the outer sphere V_outer using its charge (Q) and radius (R). V_outer = k * (Q / R)
03

Calculate the potential difference

Now, we will find the potential difference between the two spheres (V_inner and V_outer) by subtracting them. Remember, the potential difference is given by V = V_outer - V_inner. V = V_outer - V_inner V = k * (Q / R) - k * (q / r) Now, factor out k: V = k * [(Q / R) - (q / r)] We know that k equals 1 / (4 * pi * ε₀), so we can substitute that: V = (1 / (4 * pi * ε₀)) * [(Q / R) - (q / r)] Comparing this with the given options, we can see that the potential difference between the spheres is equal to option (A).

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Most popular questions from this chapter

For the system shown in figure, if the resultant force on q is zero, then \(q=\ldots \ldots \ldots\) (A) \(-2 \sqrt{2} \mathrm{Q}\) (B) \(2 \sqrt{2} \mathrm{Q}\) (C) \(2 \sqrt{3} \mathrm{Q}\) (D) \(-3 \sqrt{2} Q\)

Two point masses \(\mathrm{m}\) each carrying charge \(-\mathrm{q}\) and \(+\mathrm{q}\) are attached to the ends of a massless rigid non-conducting rod of length \(\ell\). The arrangement is placed in a uniform electric field \(\mathrm{E}\) such that the rod makes a small angle \(5^{\circ}\) with the field direction. The minimum time needed by the rod to align itself along the field is \(\ldots \ldots .\) (A) \(t=\pi \sqrt{[}(2 M \ell) /(3 q E)]\) (B) $\mathrm{t}=(\pi / 2) \sqrt{[}(\mathrm{M} \ell) /(2 \mathrm{q} \mathrm{E})]$ (D) \(t=2 \pi \sqrt{[}(M \ell / E)]\)

A charged particle of mass \(\mathrm{m}\) and charge \(q\) is released from rest in a uniform electric field \(E\). Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is ...... (A) $\left[\left(\mathrm{Eq}^{2} \mathrm{~m}\right) /\left(2 \mathrm{t}^{2}\right)\right]$ (B) $\left[\left(\mathrm{E}^{2} \mathrm{q}^{2} \mathrm{t}^{2}\right) /(2 \mathrm{~m})\right]$ (C) \(\left[\left(2 \mathrm{E}^{2} \mathrm{t}^{2}\right) /(\mathrm{qm})\right]\) (D) \([(\mathrm{Eqm}) / \mathrm{t}]\)

Three particles, each having a charge of \(10 \mu \mathrm{c}\) are placed at the corners of an equilateral triangle of side \(10 \mathrm{~cm}\). The electrostatic potential energy of the system is (Given $\left.\left[1 /\left(4 \pi \epsilon_{0}\right)\right]=9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{c}^{2}\right)$ (A) \(100 \mathrm{~J}\) (B) \(27 \mathrm{~J}\) (C) Zero (D) Infinite

A Semicircular rod is charged uniformly with a total charge \(\mathrm{Q}\) coulomb. The electric field intensity at the centre of curvature is $\ldots \ldots \ldots$ (A) \(\left[(2 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (B) \(\left[(3 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (C) \(\left[(\mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (D) \(\left[(4 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\)
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