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

A parallel plate capacitor has plate of area \(\mathrm{A}\) and separation d. It is charged to a potential difference \(\mathrm{V}_{0}\). The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is (A) $\left[\left(\mathrm{A} \in_{0} \mathrm{~V}_{0}^{2}\right) / \mathrm{d}\right]$ (B) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}^{2}\right) /(2 \mathrm{~d})\right]$ (C) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(3 \mathrm{~d})\right]$ (D) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]$

A simple pendulum consists of a small sphere of mass \(\mathrm{m}\) suspended by a thread of length \(\ell\). The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength \(\mathrm{E}\) directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is (A) $\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}-(\mathrm{q} \mathrm{E} / \mathrm{m})\\}]^{(1 / 2)}$ (B) \(\mathrm{T}=2 \pi(\ell / \mathrm{g})^{(1 / 2)}\) \(\left.\left.\left.\mathrm{m}_{\mathrm{}}\right\\}\right\\}\right]^{(1 / 2)}\) (D) \(\mathrm{T}=2 \pi[(\mathrm{m} \ell / \mathrm{qE})]^{(1 / 2)}\) (C) \(\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}+(\mathrm{qE} / \mathrm{t}\)

A simple pendulum of period \(\mathrm{T}\) has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will ...... (A) Remains equal to \(\mathrm{T}\) (B) Less than \(\mathrm{T}\) (C) Infinite (D) Greater than \(\mathrm{T}\)

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) \([(\mathrm{Eqm}) / \mathrm{t}]\)

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