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

To find the electrostatic force between the two charged spheres, we use Coulomb's Law. The formula for the magnitude of the electrostatic force (F) between two point charges ($q\_1$ and $q\_2$) separated by a distance (r) is given by:

$$F\; =\; \backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}\; \backslash cdot\; \backslash frac\{q\_1\; q\_2\}\{r^2\}$$

Where:

• $q\_1\; =\; 2\; \backslash times\; 10^\{-7\}$ C (charge on the first sphere)
• $q\_2\; =\; 3\; \backslash times\; 10^\{-7\}$ C (charge on the second sphere)
• $r\; =\; 30\; \backslash text\{\; cm\}\; =\; 0.3\; \backslash text\{\; m\}$ (distance between the spheres). It's important to convert the distance to meters for SI unit consistency.
• $\backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}\; =\; 9\; \backslash times\; 10^9\; \backslash text\{\; Nm\}^2\backslash text\{C\}^\{-2\}$ (the electrostatic constant)

Now, substitute these values into the formula:

$$F\; =\; (9\; \backslash times\; 10^9\; \backslash text\{\; Nm\}^2\backslash text\{C\}^\{-2\})\; \backslash times\; \backslash frac\{(2\; \backslash times\; 10^\{-7\}\; \backslash text\{\; C\})\; \backslash times\; (3\; \backslash times\; 10^\{-7\}\; \backslash text\{\; C\})\}\{(0.3\; \backslash text\{\; m\})^2\}$$

First, calculate the product of the charges:

$q\_1\; q\_2\; =\; (2\; \backslash times\; 10^\{-7\})\; \backslash times\; (3\; \backslash times\; 10^\{-7\})\; =\; 6\; \backslash times\; 10^\{-14\}\; \backslash text\{\; C\}^2$

Next, calculate the square of the distance:

$r^2\; =\; (0.3\; \backslash text\{\; m\})^2\; =\; 0.09\; \backslash text\{\; m\}^2$

Now, plug these back into the force equation:

$$F\; =\; (9\; \backslash times\; 10^9)\; \backslash times\; \backslash frac\{6\; \backslash times\; 10^\{-14\}\}\{0.09\}$$

Simplify the expression:

$$F\; =\; (9\; \backslash times\; 10^9)\; \backslash times\; (66.67\; \backslash times\; 10^\{-14\})$$

$$F\; =\; 600\; \backslash times\; 10^\{-5\}\; \backslash text\{\; N\}$$

$$F\; =\; 6\; \backslash times\; 10^\{-3\}\; \backslash text\{\; N\}$$

Since both charges ($q\_1$ and $q\_2$) are positive, the force between them is repulsive. Therefore, the force between the two small charged spheres is a repulsive force of magnitude $6\; \backslash times\; 10^\{-3\}$ N.

We are given the charges on two small spheres and the electrostatic force between them. We need to find the distance between them and the force on the second sphere due to the first.

Part (a): Finding the distance between the spheres

We use Coulomb's Law, which states that the magnitude of the electrostatic force (F) between two point charges ($q\_1$ and $q\_2$) separated by a distance (r) is:

$$F\; =\; \backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}\; \backslash cdot\; \backslash frac\{q\_1\; q\_2\}\{r^2\}$$

From the problem statement:

• Force, $F\; =\; 0.2$ N
• Charge on the first sphere, $q\_1\; =\; 0.4\; \backslash mu\; C\; =\; 0.4\; \backslash times\; 10^\{-6\}$ C
• Charge on the second sphere, $q\_2\; =\; -0.8\; \backslash mu\; C\; =\; -0.8\; \backslash times\; 10^\{-6\}$ C
• The electrostatic constant, $\backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}\; =\; 9\; \backslash times\; 10^9\; \backslash text\{\; Nm\}^2\backslash text\{C\}^\{-2\}$

We need to find the distance $r$. Rearranging Coulomb's Law to solve for $r^2$:

$$r^2\; =\; \backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}\; \backslash cdot\; \backslash frac\{q\_1\; q\_2\}\{F\}$$

Now, substitute the given values:

$r^2\; =\; (9\; \backslash times\; 10^9\; \backslash text\{\; Nm\}^2\backslash text\{C\}^\{-2\})\; \backslash times\; \backslash frac\{|(0.4\; \backslash times\; 10^\{-6\}\; \backslash text\{\; C\})\; \backslash times\; (-0.8\; \backslash times\; 10^\{-6\}\; \backslash text\{\; C\})|\}\{0.2\; \backslash text\{\; N\}\}$

Note: We use the absolute value of the product of charges to find the magnitude of the distance.

$r^2\; =\; (9\; \backslash times\; 10^9)\; \backslash times\; \backslash frac\{|-0.32\; \backslash times\; 10^\{-12\}|\}\{0.2\}$

$r^2\; =\; (9\; \backslash times\; 10^9)\; \backslash times\; \backslash frac\{0.32\; \backslash times\; 10^\{-12\}\}\{0.2\}$

$r^2\; =\; (9\; \backslash times\; 10^9)\; \backslash times\; (1.6\; \backslash times\; 10^\{-12\})$

$r^2\; =\; 14.4\; \backslash times\; 10^\{-3\}\; \backslash text\{\; m\}^2$

$r^2\; =\; 0.0144\; \backslash text\{\; m\}^2$

To find $r$, take the square root:

$r\; =\; \backslash sqrt\{0.0144\; \backslash text\{\; m\}^2\}\; =\; 0.12\; \backslash text\{\; m\}$

So, the distance between the two spheres is 0.12 meters.

Part (b): Force on the second sphere due to the first

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In electrostatics, this principle is known as the third law of electrostatics. The force exerted by the first sphere on the second sphere is equal in magnitude and opposite in direction to the force exerted by the second sphere on the first.

Since the force on the first sphere due to the second is given as 0.2 N, the force on the second sphere due to the first will also be 0.2 N. The charges have opposite signs ($q\_1$ is positive, $q\_2$ is negative), so this force is attractive.

Therefore, the force on the second sphere due to the first is 0.2 N (attractive).

Let's analyze the given ratio $\backslash frac\{ke^2\}\{Gm\_em\_p\}$ to check if it is dimensionless and determine its significance.

First, let's list the dimensions of each constant involved:

• $k\; =\; \backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}$ is the electrostatic constant. Its dimensions can be derived from Coulomb's Law ($F\; =\; k\; \backslash frac\{q\_1\; q\_2\}\{r^2\}$). The dimensions of force (F) are $[MLT^\{-2\}]$, charge (q) are $[IT]$ (where I is current and T is time), and distance (r) are $[L]$. So, the dimensions of k are $[\backslash frac\{MLT^\{-2\}\; \backslash cdot\; L^2\}\{(IT)^2\}]\; =\; [ML^3T^\{-2\}I^\{-2\}]$.
• $e$ is the elementary charge. Its dimensions are $[IT]$. Therefore, $e^2$ has dimensions $[I^2T^2]$.
• $G$ is the universal gravitational constant. Its dimensions are $[M^\{-1\}L^3T^\{-2\}]$.
• $m\_e$ is the mass of the electron. Its dimensions are $[M]$.
• $m\_p$ is the mass of the proton. Its dimensions are $[M]$.

Now, let's find the dimensions of the ratio $\backslash frac\{ke^2\}\{Gm\_em\_p\}$:

$\backslash [\; \backslash text\{Dimensions\; of\; ratio\}\; =\; \backslash frac\{[ML^3T^\{-2\}I^\{-2\}]\; \backslash cdot\; [I^2T^2]\}\{[M^\{-1\}L^3T^\{-2\}]\; \backslash cdot\; [M]\; \backslash cdot\; [M]\}\; \backslash ]$

Simplify the dimensions:

$\backslash [\; =\; \backslash frac\{[ML^3T^\{-2\}I^\{-2\}I^2T^2]\}\{[M^\{-1\}L^3T^\{-2\}M^2]\}\; \backslash ]$

$\backslash [\; =\; \backslash frac\{[ML^3T^0I^0]\}\{[M^\{1\}L^3T^\{-2\}]\}\; \backslash ]$

$\backslash [\; =\; [M^\{1-1\}\; L^\{3-3\}\; T^\{0-(-2)\}]\; \backslash ]$

$\backslash [\; =\; [M^0\; L^0\; T^2]\; \backslash ]$

Wait, there seems to be a mistake in the derivation of dimensions for k or in the simplification. Let's re-evaluate the dimensions of k using $k\; =\; 9\; \backslash times\; 10^9\; \backslash text\{\; Nm\}^2\backslash text\{C\}^\{-2\}$. The dimensions are $[MLT^\{-2\}][L^2][(IT)^\{-2\}]\; =\; [ML^3T^\{-2\}I^\{-2\}]$. This is correct.

Let's re-evaluate the dimensions of G: $F\; =\; G\; \backslash frac\{m\_1\; m\_2\}\{r^2\}$ => $G\; =\; \backslash frac\{F\; r^2\}\{m\_1\; m\_2\}$ => $[G]\; =\; \backslash frac\{[MLT^\{-2\}][L^2]\}\{[M][M]\}\; =\; [M^\{-1\}L^3T^\{-2\}]$. This is also correct.

Let's re-calculate the ratio dimensions:

$\backslash [\; \backslash text\{Dimensions\; of\; ratio\}\; =\; \backslash frac\{[ML^3T^\{-2\}I^\{-2\}]\; \backslash cdot\; [I^2T^2]\}\{[M^\{-1\}L^3T^\{-2\}]\; \backslash cdot\; [M]\; \backslash cdot\; [M]\}\; \backslash ]$

$\backslash [\; =\; \backslash frac\{M\; L^3\; T^0\; I^0\}\{M^\{-1\}\; L^3\; T^\{-2\}\; M^2\}\; \backslash ]$

$\backslash [\; =\; \backslash frac\{M\; L^3\}\{M^\{1\}\; L^3\; T^\{-2\}\}\; \backslash ]$

$\backslash [\; =\; M^\{1-1\}\; L^\{3-3\}\; T^\{0-(-2)\}\; =\; M^0\; L^0\; T^2\; \backslash ]$

There must be a mistake in the source or my understanding of dimensions. Let's use a common approach where charge $q$ is considered fundamental, or use SI units directly.

Let's use the definition of dimensions based on fundamental units. Charge $q$ has dimensions $[Q]$. Then $[k]\; =\; [M\; L^3\; T^\{-2\}\; Q^\{-2\}]$. $[e^2]\; =\; [Q^2]$. $[G]\; =\; [M^\{-1\}\; L^3\; T^\{-2\}]$. $[m\_e]\; =\; [M]$. $[m\_p]\; =\; [M]$.

$\backslash [\; \backslash text\{Dimensions\; of\; ratio\}\; =\; \backslash frac\{[M\; L^3\; T^\{-2\}\; Q^\{-2\}]\; \backslash cdot\; [Q^2]\}\{[M^\{-1\}\; L^3\; T^\{-2\}]\; \backslash cdot\; [M]\; \backslash cdot\; [M]\}\; \backslash ]$

$\backslash [\; =\; \backslash frac\{[M\; L^3\; T^\{-2\}]\}\{[M^\{-1\}\; L^3\; T^\{-2\}\; M^2]\}\; \backslash ]$

$\backslash [\; =\; \backslash frac\{[M\; L^3\; T^\{-2\}]\}\{[M^\{1\}\; L^3\; T^\{-2\}]\}\; \backslash ]$

$\backslash [\; =\; M^\{1-1\}\; L^\{3-3\}\; T^\{-2-(-2)\}\; =\; M^0\; L^0\; T^0\; \backslash ]$

This confirms the ratio is dimensionless. The dimensions of $k$ should be $[M\; L^3\; T^\{-2\}\; Q^\{-2\}]$ if $Q$ is a fundamental dimension, or $[M\; L^3\; T^\{-2\}\; A^\{-2\}]$ if $A$ (Ampere) is used.

Determining the value of the ratio:

We need the values of the physical constants:

• $k\; =\; 9\; \backslash times\; 10^9\; \backslash text\{\; Nm\}^2\backslash text\{C\}^\{-2\}$
• $e\; =\; 1.602\; \backslash times\; 10^\{-19\}$ C (charge of an electron)
• $G\; =\; 6.674\; \backslash times\; 10^\{-11\}\; \backslash text\{\; Nm\}^2\backslash text\{kg\}^\{-2\}$
• $m\_e\; =\; 9.109\; \backslash times\; 10^\{-31\}$ kg (mass of an electron)
• $m\_p\; =\; 1.672\; \backslash times\; 10^\{-27\}$ kg (mass of a proton)

Now, calculate the ratio:

$\backslash [\; \backslash frac\{ke^2\}\{Gm\_em\_p\}\; =\; \backslash frac\{(9\; \backslash times\; 10^9)\; \backslash times\; (1.602\; \backslash times\; 10^\{-19\})^2\}\{(6.674\; \backslash times\; 10^\{-11\})\; \backslash times\; (9.109\; \backslash times\; 10^\{-31\})\; \backslash times\; (1.672\; \backslash times\; 10^\{-27\})\}\; \backslash ]$

Calculate the numerator:

$ke^2\; =\; (9\; \backslash times\; 10^9)\; \backslash times\; (2.566\; \backslash times\; 10^\{-38\})\; =\; 23.094\; \backslash times\; 10^\{-29\}\; =\; 2.3094\; \backslash times\; 10^\{-28\}$

Calculate the denominator:

$Gm\_em\_p\; =\; (6.674\; \backslash times\; 10^\{-11\})\; \backslash times\; (9.109\; \backslash times\; 10^\{-31\})\; \backslash times\; (1.672\; \backslash times\; 10^\{-27\})$

$Gm\_em\_p\; =\; (6.674\; \backslash times\; 9.109\; \backslash times\; 1.672)\; \backslash times\; 10^\{-11\; -\; 31\; -\; 27\}$

$Gm\_em\_p\; =\; 101.75\; \backslash times\; 10^\{-69\}\; =\; 1.0175\; \backslash times\; 10^\{-67\}$

Now, divide the numerator by the denominator:

$\backslash [\; \backslash frac\{ke^2\}\{Gm\_em\_p\}\; =\; \backslash frac\{2.3094\; \backslash times\; 10^\{-28\}\}\{1.0175\; \backslash times\; 10^\{-67\}\}\; \backslash ]$

$\backslash [\; \backslash approx\; 2.2697\; \backslash times\; 10^\{39\}\; \backslash ]$

The value of the ratio is approximately $2.27\; \backslash times\; 10^\{39\}$.

Significance of the ratio:

This ratio is significant because it compares the magnitude of the electrostatic force between two elementary particles (like a proton and an electron) to the magnitude of the gravitational force between them. The ratio $\backslash frac\{F\_\{electric\}\}\{F\_\{gravitational\}\}\; \backslash approx\; 10^\{39\}$ indicates that the electrostatic force is vastly stronger than the gravitational force between charged particles. This explains why electromagnetic forces dominate at the atomic and molecular level, while gravity is more significant for large celestial bodies.