CBSE Class 12 Physics Chapter 2: Electric Potential and Capacitance NCERT Solutions

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

Charge on the first sphere, $q\_1\; =\; 2\; \backslash times\; 10^\{-7\}$ C.

Charge on the second sphere, $q\_2\; =\; 3\; \backslash times\; 10^\{-7\}$ C.

The distance between the spheres is $r\; =\; 30\; \backslash text\{\; cm\}$. To use this in the formula, we convert it to meters: $r\; =\; 30\; \backslash times\; 10^\{-2\}\; \backslash text\{\; m\}\; =\; 0.3\; \backslash text\{\; m\}$.

The electrostatic force between two point charges is given by Coulomb's law:

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

Here, $\backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}$ is Coulomb's constant, which has a value of $9\; \backslash times\; 10^\{9\}\; \backslash text\{\; Nm\}^2\backslash text\{C\}^\{-2\}$.

Now, we can substitute the given values into the formula:

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

Calculate the numerator:

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

Simplify the expression:

$$F\; =\; \backslash frac\{54\; \backslash times\; 10^\{-5\}\}\{0.09\}\; \backslash text\{\; N\}\; =\; 600\; \backslash times\; 10^\{-5\}\; \backslash text\{\; N\}\; =\; 6\; \backslash times\; 10^\{-3\}\; \backslash text\{\; N\}$$

Since both charges are positive, the force between them is repulsive.

Answer: The force between the two small charged spheres is $6\; \backslash times\; 10^\{-3\}$ N, and it is repulsive.

1. We are given the force between two charged spheres and their charges. We need to find the distance between them.

   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 force between them is $F\; =\; 0.2\; N$. Since the charges have opposite signs, the force is attractive.

   Using Coulomb's law, $F\; =\; \backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}\; \backslash cdot\; \backslash frac\{q\_1\; q\_2\}\{r^2\}$, we can rearrange to solve for $r^2$:

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

   Substitute the values:

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

   Note that we use the magnitude of the product of charges for calculating the distance.

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

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

   $r^2\; =\; 14.4\; \backslash times\; 10^\{-3\}\; \backslash text\{\; m\}^2\; =\; 0.0144\; \backslash text\{\; m\}^2$

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

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

   Answer: The distance between the two spheres is $0.12\; \backslash text\{\; m\}$ (or 12 cm).

2. According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In the context of electrostatic forces, 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 sphere.

   Therefore, the force on the second sphere due to the first sphere is equal in magnitude to the force on the first sphere due to the second sphere.

   Answer: The force on the second sphere due to the first is $0.2\; N$ (and it is attractive).

Let's analyze the dimensions of each term in the ratio $\backslash frac\{ke^2\}\{G\; m\_e\; m\_p\}$.

The constant $k$ is Coulomb's constant, $k\; =\; \backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_0\}$. The dimensions of $\backslash varepsilon\_0$ are $[M^\{-1\}\; L^\{-3\}\; T^4\; A^2]$. Therefore, the dimensions of $k$ are $[M\; L^3\; T^\{-4\}\; A^\{-2\}]$.

The charge of an electron or proton, $e$, has dimensions of electric current, $[A\; T]$. So, $e^2$ has dimensions $[A^2\; T^2]$.

The gravitational constant, $G$, has dimensions $[M^\{-1\}\; L^3\; T^\{-2\}]$.

The mass of an electron, $m\_e$, and the mass of a proton, $m\_p$, both have dimensions of mass, $[M]$. So, $m\_e\; m\_p$ has dimensions $[M^2]$.

Now, let's find the dimensions of the entire ratio:

$$\backslash left[\; \backslash frac\{ke^2\}\{G\; m\_e\; m\_p\}\; \backslash right]\; =\; \backslash frac\{[M\; L^3\; T^\{-4\}\; A^\{-2\}]\; \backslash cdot\; [A^2\; T^2]\}\{[M^\{-1\}\; L^3\; T^\{-2\}]\; \backslash cdot\; [M^2]\}$$

Simplify the dimensions:

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

Thus, the ratio $\backslash frac\{ke^2\}\{G\; m\_e\; m\_p\}$ is dimensionless.

Value of the ratio:

Looking up physical constants:

• $k\; \backslash approx\; 9\; \backslash times\; 10^9\; \backslash text\{\; Nm\}^2\backslash text\{C\}^\{-2\}$
• $e\; \backslash approx\; 1.6\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\}$
• $G\; \backslash approx\; 6.67\; \backslash times\; 10^\{-11\}\; \backslash text\{\; Nm\}^2\backslash text\{kg\}^\{-2\}$
• $m\_e\; \backslash approx\; 9.1\; \backslash times\; 10^\{-31\}\; \backslash text\{\; kg\}$
• $m\_p\; \backslash approx\; 1.67\; \backslash times\; 10^\{-27\}\; \backslash text\{\; kg\}$

Calculate $k\; e^2$:

$k\; e^2\; \backslash approx\; (9\; \backslash times\; 10^9)\; \backslash times\; (1.6\; \backslash times\; 10^\{-19\})^2\; =\; (9\; \backslash times\; 10^9)\; \backslash times\; (2.56\; \backslash times\; 10^\{-38\})\; \backslash approx\; 23.04\; \backslash times\; 10^\{-29\}\; \backslash text\{\; Nm\}^2$

Calculate $G\; m\_e\; m\_p$:

$G\; m\_e\; m\_p\; \backslash approx\; (6.67\; \backslash times\; 10^\{-11\})\; \backslash times\; (9.1\; \backslash times\; 10^\{-31\})\; \backslash times\; (1.67\; \backslash times\; 10^\{-27\})\; \backslash approx\; 101.5\; \backslash times\; 10^\{-68\}\; \backslash text\{\; Nm\}^2$

Now, calculate the ratio:

$$\backslash frac\{ke^2\}\{G\; m\_e\; m\_p\}\; \backslash approx\; \backslash frac\{23.04\; \backslash times\; 10^\{-29\}\}\{101.5\; \backslash times\; 10^\{-68\}\}\; \backslash approx\; 0.227\; \backslash times\; 10^\{39\}\; \backslash approx\; 2.27\; \backslash times\; 10^\{38\}$$

Significance of the ratio:

This ratio compares the magnitude of the electrostatic force between two protons (or an electron and a proton) to the magnitude of the gravitational force between them. The electrostatic force is significantly stronger than the gravitational force. Specifically, the ratio $\backslash frac\{ke^2\}\{G\; m\_e\; m\_p\}$ represents how many times stronger the electrostatic force is compared to the gravitational force between an electron and a proton.