Question 95: Electrostats & Capacitance

Question

Two charges '$$-q$$' each are fixed, separated by distance '$$2 d$$'. A third charge '$$q$$' of mass '$$m$$' placed at the mid-point is displaced slightly by '$$x$$' $$(x< < d)$$ perpendicular to the line joining the two fixed charges as shown in Fig. The time period of oscillation of '$$q$$' will be

Accepted Answer

To solve this problem, we need to understand a few key concepts about charge distribution and surface charge density. Let's walk through the problem step by step.
When two conductors are brought into contact, charge will redistribute between them until they reach the same electrical potential. For spheres, the surface charge density is related to the charge and the radius of the sphere. Before contact, both spheres have the same surface charge density. Let's denote this common surface charge density as $$\sigma$$.
For a sphere of radius $$R$$, the surface area $$A$$ is given by:
$$A = 4 \pi R^2$$
The charge $$Q$$ on each sphere can be expressed as:
$$Q = \sigma \cdot A = \sigma \cdot 4 \pi R^2$$
Given the radii of the two spheres, the charge on each sphere before contact is:
For the sphere with radius $$\frac{R}{2}$$:
$$Q_1 = \sigma \cdot 4 \pi \left(\frac{R}{2}\right)^2 = \sigma \cdot \pi R^2$$
For the sphere with radius $$2R$$:
$$Q_2 = \sigma \cdot 4 \pi (2R)^2 = \sigma \cdot 16 \pi R^2$$
When the two spheres are brought into contact and then separated, the total charge will be redistributed between them. The total charge is:
$$Q_{total} = Q_1 + Q_2 = \sigma \cdot \pi R^2 + \sigma \cdot 16 \pi R^2 = 17 \sigma \cdot \pi R^2$$
Since the spheres are in contact, they will have the same potential. The potential (V) for a sphere is given by:
$$V = \frac{Q}{R}$$
Let $$Q_1'$$ and $$Q_2'$$ be the charges on the spheres with radii $$\frac{R}{2}$$ and $$2R$$ respectively after they have been separated. The potential of both spheres must be equal:
$$\frac{Q_1'}{\frac{R}{2}} = \frac{Q_2'}{2R}$$
Simplifying, we get:
$$\frac{Q_1'}{R/2} = \frac{Q_2'}{2R} \implies Q_1' \cdot 2R = Q_2' \cdot \frac{R}{2} \implies Q_1' = \frac{Q_2'}{4}$$
We also know that the total charge is conserved:
$$Q_1' + Q_2' = 17 \sigma \pi R^2$$
Substituting $$Q_1' = \frac{Q_2'}{4}$$ into this equation, we get:
$$\frac{Q_2'}{4} + Q_2' = 17 \sigma \pi R^2 \implies \frac{5Q_2'}{4} = 17 \sigma \pi R^2 \implies Q_2' = \frac{68 \sigma \pi R^2}{5}$$
And for the smaller sphere:
$$Q_1' = \frac{Q_2'}{4} = \frac{68 \sigma \pi R^2}{5} \cdot \frac{1}{4} = \frac{17 \sigma \pi R^2}{5}$$
Now, we can find the new surface charge densities:
For the sphere with radius $$\frac{R}{2}$$:
$$\sigma_1' = \frac{Q_1'}{4 \pi \left(\frac{R}{2}\right)^2} = \frac{\frac{17 \sigma \pi R^2}{5}}{\pi R^2} = \frac{17\sigma}{5}$$
For the sphere with radius $$2R$$:
$$\sigma_2' = \frac{Q_2'}{4 \pi (2R)^2} = \frac{\frac{68 \sigma \pi R^2}{5}}{16 \pi R^2} = \frac{68 \sigma}{80} = \frac{17\sigma}{20}$$
Finally, the ratio of the new surface charge densities $$\sigma_1' : \sigma_2'$$ is:
$$\frac{\sigma_1'}{\sigma_2'} = \frac{\frac{17\sigma}{5}}{\frac{17\sigma}{20}} = \frac{20}{5} = 4:1$$
So, the correct answer is
Option B: 4:1
.

Answer

$$
\mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^2 \mathrm{~d}^3}{\mathrm{q}^2}}
$$

Answer

$$
\mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^3}}
$$

Answer

$$
\mathrm{T}=\sqrt{\frac{4 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^2}}
$$

Answer

$$
\mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^2}}
$$

Electrostats & Capacitance

Two charges '$-q' each are fixed, separated by distance '2 d'. A third charge 'q' of mass 'm' placed at the mid-point is displaced slightly by 'x' (x< < d) perpendicular to the line joining the two fixed charges as shown in Fig. The time period of oscillation of 'q$' will be

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