Question
Question: Two conducting spheres of radius \(R\) are placed at a large distance from each other. They are conn...
Two conducting spheres of radius R are placed at a large distance from each other. They are connected by a coil of inductance L, as shown in the figure. Sphere A is given a charge of Q and the switch ‘S’ is closed at time t=0. Find charge on sphere B as a function of time. At what time charge on B is 2Q?
Solution
The spherical capacitor of any conducting solid sphere of radius R is C=4πε0R. Where ε0 is the permittivity of free space.
The potential on the surface of a conducting sphere is V=4πε0RQ. Where Q 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 dt2d2x+ax=0. The general solution of this equation is x=Acosat+Bsinat. Now find the values of A and B 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 R.
The charge on sphere A is Q.
The sphere A and B are connected through an inductor of inductance L.
Switch S is closed at time t=0
Consider the charge on the sphere B is Q′ at a time t=t
The charge on the sphere A at the same time t=t is Q−Q′
The current through the inductor is i=dtdQ′
The potential on the surface of the sphere A, VA=4πε0RQ−Q′
The potential on the surface of the sphere B, VB=4πε0RQ′
Both the spheres have equal spherical capacitance, CA=CB=4πε0R
Sphere A and B are connected in series, The effective capacitance is C=2CA=2πε0R
Now the e.m.f. induced across the inductor is given by
E=−Ldtdi
Or VA−VB=−Ldtdi
Substitute all the required values
⇒4πε0RQ−Q′−4πε0RQ′=−Ldt2d2Q′
⇒4πε0RQ−2Q′=−Ldt2d2Q′
Further simplify to get a differential equation.
⇒dt2d2Q′+2πε0RL1(2Q−Q′)=0
We know that the angular frequency ω=LC1
Or ω=2πε0RL1
⇒dt2d2Q′+ω(2Q−Q′)=0
Let q=2Q−Q′, Then dt2d2q=−dt2d2Q′
The above equation becomes
⇒dt2d2q−ωq=0 …… (1)
The general Solution of the above differential equation is given by
q=Acosωt−Bsinωt …… (2)
Now we have to find the values of A and B.
At time t=0, Q′=0, So q=Q
Q=Acos0−Bsin0
⇒A=Q
Now i=dtdQ′=−dtdq
Or i=−(−Asinωt−Bcosωt)
Or i=Asinωt+Bcosωt
At time t=0, i=0, We got
0=Asin0+Bcos0
⇒B=0
The equation (2) becomes
q=Qcosωt
⇒2Q−Q′=Qcosωt
⇒Q′=2Q(1−cosωt) …… (3)
Hence, the charge on the surface of the sphere B at a time t=t is 2Q(1−cosωt).
Now we have to find the time when the charge on sphere B is 2Q.
i.e., From equation (3), 1−cosωt=1
⇒cosωt=0
⇒cosωt=cos2π
Or ωt=2π
Now substitute the value ω=2πε0RL1, We got
⇒t=2π2πε0RL
Hence, at the time t=2π2πε0RL, the charge on the surface of the sphere B will be 2Q.
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=2πLC1.