Class 12 Physics Chapter 2 Electrostatics Potential and Capacitance

Key Concept: Basic Concept

b) $U = -p \cdot E \cdot \cos(\theta)$

[Solution Description] The potential energy $U(\theta)$ of a dipole in a uniform electric field is given by the formula $U = -p \cdot E \cdot \cos(\theta)$, where $p$ is the dipole moment, $E$ is the electric field strength, and $\theta$ is the angle between $p$ and $E$. This formula arises because a dipole tends to align with the electric field, minimizing its potential energy.

Key Concept: Conservative Force

d) Because the field does work that can be fully recovered without loss.
[Solution Description] Potential energy is meaningful in an electrostatic field because it is a conservative field. This means that the work done on a charge moving within the field is stored as potential energy, allowing for energy conservation analysis.

Key Concept: Conductors and Fields

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] In an electrostatic equilibrium, conductors have no net movement of charge, leading to the cancellation of any internal electric fields. This happens because free charges redistribute on the surface until the internal field is zero. Since both the assertion and reason are true and the redistribution of charges effectively explains why the internal field becomes zero, the reason provided correctly explains the assertion.

Key Concept: Series Formula

b) $1/C = 1/C_1 + 1/C_2 + 1/C_3$

[Solution Description] For capacitors connected in series, the reciprocal of the total or effective capacitance $C$ can be found by summing the reciprocals of the individual capacitances. The formula is given by: $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + … + \frac{1}{C_n}$.

Key Concept: Equipotential for Dipole

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description] A dipole consists of two equal and opposite charges separated by some distance. The equipotential surfaces around a dipole are neither perfectly spherical nor symmetrical as they are with a single point charge. Instead, the potential is constant along curved surfaces that reflect the influence of both charges, resulting in a more complex shape. This complexity arises because the potential due to each charge combines to form these surfaces. On the other hand, for a single point charge, the equipotential surfaces are indeed spherical due to the uniform radial symmetry from the center of the charge. Thus, while both statements correctly describe different scenarios, Reason does not explain Assertion since it pertains to a separate situation involving a single point charge.

Key Concept: Advanced Voltage Calculation, Energy Stored

c) 1.15V
[Solution Description] First, find the equivalent capacitance in series using:

$$
\frac{1}{C_{eq}} = \frac{1}{1 \, \mu F} + \frac{1}{5 \, \mu F} + \frac{1}{10 \, \mu F}
= \frac{10}{10} + \frac{2}{10} + \frac{1}{10}
= \frac{13}{10}
$$

Therefore,

$C_{eq} = \frac{10}{13} \, \mu F$

The charge $Q$ on each capacitor can be calculated as:

$Q = C_{eq} \times V = \frac{10}{13} \times 15 = \frac{150}{13} \, \mu C$

The voltage across the 10 $\mu F$ capacitor is given by:

$V_3 = \frac{Q}{C_3} = \frac{\frac{150}{13}}{10} = \frac{15}{13} \approx 1.15 \, V$

So the voltage across the 10 $\mu F$ capacitor is approximately 1.15V.

Key Concept: Basic Concept

d) $qV(r)$

[Solution Description] The potential energy $U$ of a charge $q$ in an external electric field at a point where the potential is $V(r)$ is given by the formula $U = qV(r)$.

Key Concept: Basic Formula

b) $C = \frac{\epsilon_0 K A}{d}$

[Solution Description] The capacitance of a parallel plate capacitor with a dielectric inserted is given by the formula $C = \frac{\epsilon_0 K A}{d}$. Here, $\epsilon_0$ is the permittivity of free space, K is the dielectric constant, A is the area of one of the plates, and d is the separation between the plates.

Key Concept: Basic Formula Application

d) 135000 V

[Solution Description] Using the formula for potential due to a single charge, $$V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$$. Substituting the given values:

$V = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \times \frac{3 \times 10^{-6} \, \text{C}}{0.2 \, \text{m}}$

Performing the calculation,

$V = 9 \times 10^9 \times 15 \times 10^{-6} = 135000 \, \text{V}$

Hence, the potential at point P is 135000 V.

Key Concept: Dipole in Field, Variation of Dipole Potential with Distance

b) $pE$
[Solution Description] The potential energy $U$ of a dipole in an external field is $U = -pE \cos \theta$, where $\theta$ is the angle between the dipole moment and the electric field. Initially, $\theta = 0^\circ$, so $U_{\text{initial}} = -pE $.

Rotating the dipole by $90^\circ$, changes $\theta$ to $90^\circ$, leading to:
$ U_{\text{final}} = -pE \cos 90^\circ = 0 $

The change in potential energy $\Delta U$ is:
$\Delta U = U_{\text{final}} – U_{\text{initial}} = 0 – (-pE) = pE$

Therefore, option b) corresponds to the computed result.

Key Concept: Induced vs Permanent Dipoles

c) Induced dipoles are created by an external electric field
[Solution Description] Induced dipoles occur when a non-polar molecule experiences an external electric field, causing a temporary shift in charge distribution that induces a dipole moment. Permanent dipoles are inherent to polar molecules due to their asymmetric charge distribution. In an external electric field, these permanent dipoles align with the field. Therefore, induced dipoles are not inherently aligned but rather result from an external influence, whereas permanent dipoles can naturally align with an applied field.

Key Concept: Thermal Agitation

c) It reduces the alignment due to increased randomness.
[Solution Description] When an external electric field is applied to polar molecules, they tend to align with the field. However, thermal agitation acts as a randomizing force that disrupts this alignment by causing constant rotational motion of the molecules. As temperature increases, the degree of alignment decreases because increased kinetic energy from heat causes more random movement.

Key Concept: Dielectric Constant Definition

c) $\frac{\varepsilon}{\varepsilon_0}$
[Solution Description]
The dielectric constant is defined as the ratio of the permittivity of the substance ($\varepsilon$) to the permittivity of free space ($\varepsilon_0$):
$K = \frac{\varepsilon}{\varepsilon_0}$

Key Concept: Charge Configuration

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Since electrostatic forces are conservative, the work done in bringing a charge from infinity to a point does not depend on the path taken but only on the initial and final positions. This characteristic allows us to consider the potential as a scalar quantity that can be summed up algebraically for each contributing charge. Thus, the assertion holds true because we simply add each charge’s contribution to find the total potential at a point. However, while the reason provided is true — indicating the conservative nature of electrostatic forces — it is not directly explaining why the potential is the algebraic sum of contributions from individual charges.

Key Concept: Basic Formula

b) $\frac{1}{C} = \frac{1}{C1} + \frac{1}{C2}$
[Solution Description]
The formula for finding the effective capacitance $C$ when two capacitors $C1$ and $C2$ are connected in series is given by:
$\frac{1}{C} = \frac{1}{C1} + \frac{1}{C2}$
This means that the reciprocal of the total capacitance is equal to the sum of the reciprocals of the individual capacitances.

Key Concept: Definition Recall

c) Electric field is the gradient of potential

[Solution Description] An electric field is defined as the gradient of the potential; hence it is related to how the potential changes spatially.

Key Concept: Unit Identification

c) Farad

[Solution Description] The unit of capacitance in the International System of Units (SI) is the farad, abbreviated as F. It measures a capacitor’s ability to store charge per unit voltage.

Key Concept: Basic Formula

c) 20 F
[Solution Description]
To find the effective capacitance for capacitors in parallel, we use the formula $C = C_1 + C_2 + C_3$. Substituting the given values:

$C = 4\,F + 6\,F + 10\,F = 20\,F$

Therefore, the effective capacitance is 20 F.

Key Concept: Dielectric Strength

c) The maximum electric field a dielectric medium can withstand
[Solution Description] Dielectric strength is defined as the maximum electric field that a dielectric material can withstand without undergoing electrical breakdown. It is expressed in volts per meter (V/m). Therefore, a higher dielectric strength indicates better insulation properties.

Key Concept: Simple Calculation

b) -15 J

[Solution Description] The potential energy $U$ of a dipole in a uniform electric field is calculated using the formula $U = -p \cdot E$. Substituting the given values:

$U = -(3 \, \text{Cm}) \times (5 \, \text{N/C}) = -15 \, \text{J}.$

So, the potential energy is $-15 \, \text{J}$.

Key Concept: Theoretical Implications, Conceptual Analysis

b) $-\frac{1}{2}U_i$
[Solution Description]
Initially, the potential energy is given by:
$U_i = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$

When the distance is doubled to $2r$, the new potential energy becomes:
$U_f = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{2r} = \frac{1}{2}\left(\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}\right)$

The change in potential energy is:
$\Delta U = U_f – U_i = \frac{1}{2}U_i – U_i = -\frac{1}{2}U_i$

Therefore, the change in potential energy is $-\frac{1}{2}U_i$.

Key Concept: Electrostatic Shielding

c) Electrostatic shielding.
[Solution Description] Electrostatic shielding ensures that the electric field inside the cavity of a Van de Graaff generator is zero. This is because charges reside on the outer surface, preventing any internal electric field.

Key Concept: Unit Recognition

b) Joules per cubic meter
[Solution Description] Energy density is defined as energy stored per unit volume, and its SI unit is Joules per cubic meter $(J/m^3).$

Key Concept: Equipotential and Energy, Advanced Field Orientation

c) The entire space is an equipotential surface; no energy is required to move the charge.
[Solution Description] Inside a conductor in electrostatic equilibrium, the electric potential is uniform due to the redistribution of surface charges. Therefore, any point inside the cavity of the conductor has the same potential as any other point within the cavity.

For this configuration, the entire volume inside the conducting shell acts as an equipotential region because there is no net electric field in the conductor’s interior once the charges have redistributed themselves (the charges on the inner surface of the shell neutralize the effect of the central charge).

Hence, moving a test charge between any two points within the shell requires no work, as there is no difference in electric potential.

Key Concept: Conceptual Understanding

b) Doubles
[Solution Description]
The relationship between charge (Q), capacitance (C), and potential difference (V) is given by:

$V = \frac{Q}{C}$

If capacitance C is halved, new capacitance becomes $\frac{C}{2}$. Therefore, the new potential difference V’ is:

$V’ = \frac{Q}{(C/2)} = \frac{2Q}{C} = 2V$

Hence, the potential difference doubles when the capacitance is halved.

Key Concept: Dielectric Effect

c) It decreases
[Solution Description]
Initially, without the dielectric, the electric field $\left(E_0\right)$ between the plates is given by $E_0 = V/d$, where V is the potential difference and d is the separation.
When a dielectric with a dielectric constant K is introduced, the electric field within the dielectric reduces because the dielectric polarizes in response to the applied field, creating an induced charge that opposes the external field.
The new electric field $E$ becomes $E_0/K$. Since $K > 1$, we have $E < E_0$. Therefore, the electric field decreases.

Key Concept: Electrostatic Shielding

d) It ensures the electric field inside the cavity is zero
[Solution Description] Electrostatic shielding ensures that the electric field inside a conductor’s cavity is zero. This means any sensitive instruments placed within this cavity are protected from external electric fields.

Key Concept: Three Charges System

c) -246 J
[Solution Description] The potential energy $U$ for three charges is given by
$U = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right).$

Substituting the values:
$U = \frac{1}{4\pi\varepsilon_0} \left( \frac{(2 \times 10^{-6}) (-3 \times 10^{-6})}{0.2} + \frac{(2 \times 10^{-6}) (4 \times 10^{-6})}{0.3} + \frac{(-3 \times 10^{-6}) (4 \times 10^{-6})}{0.5} \right) $
$= \frac{1}{4\pi\varepsilon_0} \left( -30 \times 10^{-12} + \frac{80 \times 10^{-12}}{3} – 24 \times 10^{-12} \right) $
$= \frac{1}{4\pi\varepsilon_0} \left( -30 + 26.67 – 24 \right) \times 10^{-12} )$
$= \frac{-27.33 \times 10^{-12}}{4\pi\varepsilon_0}.$

Considering $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, we have:

$U = 9 \times 10^9 \times -27.33 \times 10^{-12} $
$= -246 \, \text{J}.$
Therefore, the potential energy of the system is approximately -246 J.

Key Concept: Basic Formula

c) 0.0405 J

[Solution Description]

To find the potential energy, we use the formula:

$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$

where $q_1 = 6 \times 10^{-6} \, C$, $q_2 = 3 \times 10^{-6} \, C$, and $r_{12} = 4 \, m$. The value of $\frac{1}{4\pi\epsilon_0}$ is $9 \times 10^9 \, N \cdot m^2/C^2$.

Substituting these values into the formula gives:

$U = 9 \times 10^9 \times \frac{(6 \times 10^{-6}) \times (3 \times 10^{-6})}{4}$

$U = 9 \times 10^9 \times \frac{18 \times 10^{-12}}{4}$

$U = 9 \times 10^9 \times 4.5 \times 10^{-12}$

$U = 40.5 \times 10^{-3} \, J$

$U = 0.0405 \, J$

So, the potential energy is $0.0405\, J$.

Key Concept: Units of Energy

a) Joule
[Solution Description] The energy stored in a capacitor is measured in units of joules (J). This is a standard unit for energy in the International System of Units (SI).

Key Concept: Conceptual Understanding

d) It increases
[Solution Description]
In a parallel connection, the total capacitance $C_{\text{total}}$ is the sum of the individual capacitances. So if two capacitors $C_1$ and $C_2$ are connected in parallel, their total capacitance is given by:
$C_{\text{total}} = C_1 + C_2.$

Hence, the total capacitance increases when capacitors are connected in parallel.