91影视

a)

The charge in uniformly distributed, therefore the total charge is:

\(\begin{aligned}Q = \sigma {A_A}\\ = \sigma ({A_ \circ } - {A_i})\\ = \sigma \pi ({R_2}^2 - {R_1}^2)\end{aligned}\)

Therefore, the total charge is \(\sigma \pi ({R_2}^2 - {R_1}^2)\)

b)

The electric field of the annuls is similar to the dick, but instead the radius is from \(0 \to R\)instead from \({R_1} \to {R_2}\).

Therefore,

\(\begin{aligned}E = {E_x}\\{E_x} = \frac{{\sigma x}}{{4{\varepsilon _ \circ }}}\int_{{R_1}}^{{R_2}} {\frac{{2rdr}}{{{{({x^2} + {r^2})}^{\frac{3}{2}}}}}} \\ = \frac{{\sigma x}}{{2{\varepsilon _ \circ }}}\left( { - \frac{1}{{{{({x^2} - {r^2})}^{\frac{1}{2}}}}}} \right)_{{R_2}}^{{R_1}}\\ = \frac{{\sigma x}}{{2{\varepsilon _ \circ }}}\left( { - \frac{1}{{{{({x^2} + {R_2}^2)}^{\frac{1}{2}}}}} + \frac{1}{{{{({x^2} + {R_1}^2)}^{\frac{1}{2}}}}}} \right)\end{aligned}\)

Hence the required value is \(E = \frac{{\sigma x}}{{2{\varepsilon _ \circ }}}\left( { - \frac{1}{{{{({x^2} + {R_2}^2)}^{\frac{1}{2}}}}} + \frac{1}{{{{({x^2} + {R_1}^2)}^{\frac{1}{2}}}}}} \right)\)

c)

If \(x < < {R_1}\)and \(x < < {R_2}\)then the above equation can be reduced to

\(E = \frac{{\sigma x}}{{2{\varepsilon _ \circ }}}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\)

Here, \(x\)is a variable and the electric field is proportional to it at a very close distance.

d)

The restoring force according to simple harmonic motion is given by: \(F = - kd{\rm{ }} \ldots \ldots \ldots .\left( i \right)\)

It is given that the equilibrium distance\(x{\rm{ }} = {\rm{ }} - 0.01{R_1}\), that is \(x < < {R_1}\).

Therefore, the electric force at equilibrium position is

\(kd = \frac{{\sigma qx}}{{2{\varepsilon _{}}}}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)..........(ii)\)

But it is also given that \(d = x\)

Therefore \(k\) is given by:

\(k = \frac{{\sigma q}}{{2{\varepsilon _ \circ }}}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)..........(iii)\)

Therefore, the frequency is given by:

\(\begin{aligned}f = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} \\ = \frac{1}{{2\pi }}\sqrt {\frac{{\sigma q}}{{2m{\varepsilon _ \circ }}}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)} \end{aligned}\)

Hence the frequency is \(\frac{1}{{2\pi }}\sqrt {\frac{{\sigma q}}{{2m{\varepsilon _ \circ }}}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)} \)