CBSE Class 12 Physics Chapter 1: Electric Charges and Fields NCERT Solutions

In Millikan's oil drop experiment, the oil drop is held stationary when the upward electric force acting on it exactly balances its downward gravitational force (weight).

Given:

• Number of excess electrons, $n\; =\; 12$
• Electric field intensity, $E\; =\; 2.55\; \backslash times\; 10^4\; \backslash text\{\; N\; C\}^\{-1\}$
• Density of oil, $\backslash rho\; =\; 1.26\; \backslash text\{\; g/cm\}^3$
• Acceleration due to gravity, $g\; =\; 9.81\; \backslash text\{\; m\; s\}^\{-2\}$
• Charge of an electron, $e\; =\; 1.60\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\}$

First, we convert the density of oil to SI units:

$\backslash rho\; =\; 1.26\; \backslash text\{\; g/cm\}^3\; =\; 1.26\; \backslash times\; \backslash frac\{10^\{-3\}\; \backslash text\{\; kg\}\}\{(10^\{-2\}\; \backslash text\{\; m\})^3\}\; =\; 1.26\; \backslash times\; \backslash frac\{10^\{-3\}\}\{10^\{-6\}\}\; \backslash text\{\; kg/m\}^3\; =\; 1.26\; \backslash times\; 10^3\; \backslash text\{\; kg/m\}^3$

The net charge on the oil drop is $q\; =\; ne\; =\; 12\; \backslash times\; 1.60\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\}$.

The electric force on the drop is $F\_E\; =\; qE\; =\; (ne)E$.

The weight of the oil drop is $W\; =\; mg$. The mass $m$ can be expressed using the density $\backslash rho$ and the volume of the drop $V$. Assuming the drop is spherical with radius $r$, its volume is $V\; =\; \backslash frac\{4\}\{3\}\backslash pi\; r^3$. Therefore, $m\; =\; \backslash rho\; V\; =\; \backslash rho\; \backslash times\; \backslash frac\{4\}\{3\}\backslash pi\; r^3$.

So, the weight is $W\; =\; \backslash left(\backslash frac\{4\}\{3\}\backslash pi\; r^3\; \backslash rho\backslash right)\; g$.

For the drop to be stationary, $F\_E\; =\; W$:

$qE\; =\; mg$

$(ne)E\; =\; \backslash left(\backslash frac\{4\}\{3\}\backslash pi\; r^3\; \backslash rho\backslash right)\; g$

Now, we rearrange the equation to solve for the radius $r$:

$$r^3\; =\; \backslash frac\{3(ne)E\}\{4\backslash pi\; \backslash rho\; g\}$$

Substitute the given values:

$$r^3\; =\; \backslash frac\{3\; \backslash times\; (12\; \backslash times\; 1.60\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\})\; \backslash times\; (2.55\; \backslash times\; 10^4\; \backslash text\{\; N\; C\}^\{-1\})\}\{4\; \backslash pi\; \backslash times\; (1.26\; \backslash times\; 10^3\; \backslash text\{\; kg/m\}^3)\; \backslash times\; (9.81\; \backslash text\{\; m\; s\}^\{-2\})\}$$

Using $\backslash pi\; \backslash approx\; 3.14159$:

$$r^3\; =\; \backslash frac\{3\; \backslash times\; 1.92\; \backslash times\; 10^\{-18\}\; \backslash times\; 2.55\; \backslash times\; 10^4\}\{4\; \backslash times\; 3.14159\; \backslash times\; 1.26\; \backslash times\; 10^3\; \backslash times\; 9.81\}$$

$$r^3\; =\; \backslash frac\{1.4688\; \backslash times\; 10^\{-13\}\}\{1.5457\; \backslash times\; 10^5\}\; \backslash approx\; 9.502\; \backslash times\; 10^\{-19\}\; \backslash text\{\; m\}^3$$

Now, take the cube root to find $r$:

$$r\; =\; \backslash left(9.502\; \backslash times\; 10^\{-19\}\backslash right)^\{\backslash frac\{1\}\{3\}\}\; \backslash text\{\; m\}$$

$$r\; \backslash approx\; 9.83\; \backslash times\; 10^\{-7\}\; \backslash text\{\; m\}$$

The radius of the drop is approximately $9.83\; \backslash times\; 10^\{-7\}$ meters.

Electrostatic field lines have specific properties that must be obeyed. Let's analyze each option:

• Option (a): This figure shows electric field lines inside a conductor. In electrostatic equilibrium, the electric field inside a conductor must be zero. If there were field lines inside, it would imply a non-zero field, which contradicts the condition of electrostatic equilibrium. Therefore, this cannot represent electrostatic field lines in equilibrium.
• Option (b): This figure shows field lines originating from positive charges and terminating on negative charges. This is consistent with the property that electric field lines originate from positive charges and terminate on negative charges.
• Option (c): This figure shows field lines originating from positive charges and extending outwards. This is also consistent with the property that electric field lines originate from positive charges and extend to infinity if there are no negative charges nearby.

Based on the properties of electrostatic field lines, specifically that the electric field inside a conductor in electrostatic equilibrium is zero, the representation in figure (a) is not possible.

Answer: The curve shown in figure (a) cannot possibly represent electrostatic field lines inside a conductor in equilibrium.