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Question: Two conducting spheres of radius \(R\) are placed at a large distance from each other. They are conn...

Two conducting spheres of radius RR are placed at a large distance from each other. They are connected by a coil of inductance LL, as shown in the figure. Sphere A is given a charge of QQ and the switch ‘S’ is closed at time t=0t = 0. Find charge on sphere B as a function of time. At what time charge on B is Q2\dfrac{Q}{2}?

Explanation

Solution

The spherical capacitor of any conducting solid sphere of radius RR is C=4πε0RC = 4\pi {\varepsilon _0}R. Where ε0{\varepsilon _0} is the permittivity of free space.
The potential on the surface of a conducting sphere is V=Q4πε0RV = \dfrac{Q}{{4\pi {\varepsilon _0}R}}. Where QQ is the total charge on the surface of the sphere.
Find the potential difference across the inductor. Then you will get a familiar differential equation like d2xdt2+ax=0\dfrac{{{d^2}x}}{{d{t^2}}} + ax = 0. The general solution of this equation is x=Acosat+Bsinatx = A\cos at + B\sin at. Now find the values of AA and BB by the application of the initial condition.

Complete step by step answer:
It is given that the radius of the two conducting spheres A and B is RR.
The charge on sphere A is QQ.
The sphere A and B are connected through an inductor of inductance LL.
Switch S is closed at time t=0t = 0
Consider the charge on the sphere B is QQ' at a time t=tt = t
The charge on the sphere A at the same time t=tt = t is QQQ - Q'
The current through the inductor is i=dQdti = \dfrac{{dQ'}}{{dt}}
The potential on the surface of the sphere A, VA=QQ4πε0R{V_A} = \dfrac{{Q - Q'}}{{4\pi {\varepsilon _0}R}}
The potential on the surface of the sphere B, VB=Q4πε0R{V_B} = \dfrac{{Q'}}{{4\pi {\varepsilon _0}R}}
Both the spheres have equal spherical capacitance, CA=CB=4πε0R{C_A} = {C_B} = 4\pi {\varepsilon _0}R
Sphere A and B are connected in series, The effective capacitance is C=CA2=2πε0RC = \dfrac{{{C_A}}}{2} = 2\pi {\varepsilon _0}R
Now the e.m.f. induced across the inductor is given by
E=LdidtE = - L\dfrac{{di}}{{dt}}
Or VAVB=Ldidt{V_A} - {V_B} = - L\dfrac{{di}}{{dt}}
Substitute all the required values
QQ4πε0RQ4πε0R=Ld2Qdt2\Rightarrow \dfrac{{Q - Q'}}{{4\pi {\varepsilon _0}R}} - \dfrac{{Q'}}{{4\pi {\varepsilon _0}R}} = - L\dfrac{{{d^2}Q'}}{{d{t^2}}}
Q2Q4πε0R=Ld2Qdt2\Rightarrow \dfrac{{Q - 2Q'}}{{4\pi {\varepsilon _0}R}} = - L\dfrac{{{d^2}Q'}}{{d{t^2}}}
Further simplify to get a differential equation.
d2Qdt2+12πε0RL(Q2Q)=0\Rightarrow \dfrac{{{d^2}Q'}}{{d{t^2}}} + \dfrac{1}{{2\pi {\varepsilon _0}RL}}\left( {\dfrac{Q}{2} - Q'} \right) = 0
We know that the angular frequency ω=1LC\omega = \sqrt {\dfrac{1}{{LC}}}
Or ω=12πε0RL\omega = \sqrt {\dfrac{1}{{2\pi {\varepsilon _0}RL}}}
d2Qdt2+ω(Q2Q)=0\Rightarrow \dfrac{{{d^2}Q'}}{{d{t^2}}} + \omega \left( {\dfrac{Q}{2} - Q'} \right) = 0
Let q=Q2Qq = \dfrac{Q}{2} - Q', Then d2qdt2=d2Qdt2\dfrac{{{d^2}q}}{{d{t^2}}} = - \dfrac{{{d^2}Q'}}{{d{t^2}}}
The above equation becomes
d2qdt2ωq=0\Rightarrow \dfrac{{{d^2}q}}{{d{t^2}}} - \omega q = 0 …… (1)
The general Solution of the above differential equation is given by
q=AcosωtBsinωtq = A\cos \omega t - B\sin \omega t …… (2)
Now we have to find the values of AA and BB.
At time t=0t = 0, Q=0Q' = 0, So q=Qq = Q
Q=Acos0Bsin0Q = A\cos 0 - B\sin 0
A=Q\Rightarrow A = Q
Now i=dQdt=dqdti = \dfrac{{dQ'}}{{dt}} = - \dfrac{{dq}}{{dt}}
Or i=(AsinωtBcosωt)i = - \left( { - A\sin \omega t - B\cos \omega t} \right)
Or i=Asinωt+Bcosωti = A\sin \omega t + B\cos \omega t
At time t=0t = 0, i=0i = 0, We got
0=Asin0+Bcos00 = A\sin 0 + B\cos 0
B=0\Rightarrow B = 0
The equation (2) becomes
q=Qcosωtq = Q\cos \omega t
Q2Q=Qcosωt\Rightarrow \dfrac{Q}{2} - Q' = Q\cos \omega t
Q=Q2(1cosωt)\Rightarrow Q' = \dfrac{Q}{2}\left( {1 - \cos \omega t} \right) …… (3)
Hence, the charge on the surface of the sphere B at a time t=tt = t is Q2(1cosωt)\dfrac{Q}{2}\left( {1 - \cos \omega t} \right).
Now we have to find the time when the charge on sphere B is Q2\dfrac{Q}{2}.
i.e., From equation (3), 1cosωt=11 - \cos \omega t = 1
cosωt=0\Rightarrow \cos \omega t = 0
cosωt=cosπ2\Rightarrow \cos \omega t = \cos \dfrac{\pi }{2}
Or ωt=π2\omega t = \dfrac{\pi }{2}
Now substitute the value ω=12πε0RL\omega = \sqrt {\dfrac{1}{{2\pi {\varepsilon _0}RL}}} , We got
t=π22πε0RL\Rightarrow t = \dfrac{\pi }{2}\sqrt {2\pi {\varepsilon _0}RL}
Hence, at the time t=π22πε0RLt = \dfrac{\pi }{2}\sqrt {2\pi {\varepsilon _0}RL} , the charge on the surface of the sphere B will be Q2\dfrac{Q}{2}.

Note:
The given circuit diagram is a LC oscillations circuit. Where the two conducting spheres act as spherical capacitors. When a charged capacitor is allowed to discharge through an inductor, electrical oscillations are produced. These oscillations are called LC oscillations.
The frequency of oscillating charge is given by f=12πLCf = \dfrac{1}{{2\pi \sqrt {LC} }}.