Continuous Charge Distribution: Physics class 12th Numerical

 Continuous Charge Distribution

    There are three types of continuous distribution 

1.    Linear Charge Distribution

2.    Surface Charge Distribution

3.    Volume Charge Distribution

Linear Charge Distribution

    When the charge is distributed uniformly along a line then it is a linear charge distribution.

Linear Charge Density

    The ratio of charge dq and dl is called linear charge density. Where dq is charge on extremely small length and dl is very small length.

We denote the linear charge density by `\lamda`

S.I. unit of linear charge density is Coulomb/meter.

Dimension of linear charge density is `[M^0L^{-1}T^1A^1]`

`\lamda = \frac{dq}{dl}`

`dq = \lamda dl`

    If the length of the wire is `l` and total charge q is uniformly distributed on the wire, then linear charge density

 `\lamda = \frac{q}{l}` 

Surface Charge Distribution

    When the charge is distributed uniformly along a surface then it is a surface charge distribution.

Surface Charge Density

    The ratio of charge dq and dA is called surface charge density. Where dq is the charge on an extremely small area and dA is very small area.

We denote the surface charge density by `\sigma`

S.I. unit of surface charge density is Coulomb/`meter^2`.

Dimension of surface charge density is `[M^0L^{-2}T^1A^1]`

`\lamda = \frac{dq}{dA}`

`dq = \sigma dA`

    If the area of the surface is A and total charge q is uniformly distributed on the surface, then surface charge density

 `\sigma = \frac{q}{A}` 

Volume Charge Distribution

    When the charge is distributed uniformly along volume then it is a volume charge distribution.

Volume Charge Density

    The ratio of charge dq and dA is called volume charge density. Where dq is a charge on an extremely small area and dA is a very small area.

We denote the volume charge density by `\rho`

S.I. unit of volume charge density is Coulomb/`meter^3`.

Dimension of volume charge density is `[M^0L^{-3}T^1A^1]`

`\lamda = \frac{dq}{dV}`

`dq = \rho dV`

    If the area of the volume is V and total charge q is uniformly distributed on volume, then volume charge density

 `\rho = \frac{q}{V}` 

Related Questions

1. What is the continuous charge distribution?
2. What is the difference between continuous and discrete charge distribution
3. What is charge distribution mean?
4. What are the three types of continuous charge distribution?
5. What are the dimensions of linear charge density?
6. What are the dimensions of area charge density?
7. What are the dimensions of volume charge density?
8. What are  the units of linear charge density?
9. What are the units of area charge density?
10. What are the units of volume charge density?

Question 1

A wire of length 2 meters carries a total charge of 6 Coulombs. Calculate the linear charge density.

We can use this formula for linear charge density

`\lambda = \frac{Q}{L}`

Where,

`\lamda =` linear charge density

Q = Total Charge, and

L = Length of wire

According to question

Q = 6 Coulombs

L = 2 meters

Then,

`\lambda = \frac{Q}{L}`

`\lambda = \frac{6}{2}\frac{C}{m}`

`\lambda = 3 \frac{C}{m}`

So, the linear charge density is 3 `\frac{C}{m}`

Question 2

    If a wire has a linear charge density of 10 μC/m, and its length is 5 meters, what is the total charge on the wire?

We can use this formula for linear charge density

`\lambda = \frac{Q}{L}`

`Q = \lambda \times L`

Where,

`\lamda =` linear charge density

Q = Total Charge, and

L = Length of wire

According to question

`\lambda` = 10 `\frac{\mu C}{m}` Coulombs/meter

L = 5 meters

Then,

`Q = 10 \times 5`

`Q = 50 \mu` Coulombs

Question 3

    A surface with an area of 0.5 m² has a total charge of 16 μC distributed uniformly. Find the surface charge density.

Solution :

We can use the formula for surface charge density

`\sigma = \frac{Q}{A}`

Where,

`\sigma = ` Surface charge density,

Q = Total Charge, and

A = Area

According to Question

Q = 16 `\mu` Coulombs

A = 0.5 `m^2`

Then,

`\sigma = \frac{Q}{A}`

`\sigma = \frac{16}{0.5}\frac{C}{m^2}`

`\sigma = \frac{160}{5}\frac{C}{m^2}`

`\sigma = 32 \frac{\mu C}{m^2}`

So, the surface charge density is 32 `\frac{\mu C}{m^2}`

Question 4

    The surface charge density of a charged plate is 2 μC/m². If the area of the plate is 3 m², calculate the total charge on the plate.

We can use this formula for surface charge density

`\sigma = \frac{Q}{A}`

`Q = \sigma \times A`

Where,

`\sigma =` Surface charge density

Q = Total Charge, and

A = Area

According to question

`\sigma` = 2 \mu Coulombs/mete`r^2`

`A = 3 m^2` 

Then,

`Q = \sigma \times A`

`Q = 2 \times 3`

`Q = 6` `\mu C`

Question 5

    A sphere has a volume of 1 m³ and a total charge of 12 μC uniformly distributed within it. Determine the volume charge density.

Solution :

We can use the formula for volume charge density

`\rho = \frac{Q}{V}`

Where,

`\rho = ` Volume charge density,

Q = Total Charge, and

V = Volume

According to Question

Q = 12 `\mu C`

A = 1`m^3`

Then,

`\rho = \frac{Q}{V}`

`\rho = \frac{12}{1}\frac{\mu C}{m^3}`

`\rho = 12\frac{\mu C}{m^3}`

So, the volume charge density is 12 `\frac{\mu C}{m^3}`

Question 6

    If the volume charge density of a charged cube is 5 μC/m³ and its volume is 0.2 m³, what is the total charge inside the cube?

We can use this formula for surface charge density

`\rho = \frac{Q}{V}`

`Q = \rho \times V`

Where,

`\rho =` Volume charge density

Q = Total Charge, and

V = Volume

According to question

`\rho = 5 \frac{\mu C}{m^3}`

`V = 0.2 m^3` 

Then,

`Q = 5 \times 0.2`

`Q = 1 \mu C`

Question 7

    A wire has a linear charge density of 8 nC/m and a length of 4 meters. Calculate the charge on a segment of length 1.5 meters.

We can use this formula for linear charge density

`\lambda = \frac{Q}{L}`

`Q = \lambda \times L`

Where,

`\lamda =` linear charge density

Q = Total Charge, and

L = Length of wire

According to question

`\lambda` = 8 `\frac{nC}{m}`

L = 1.5 meters

Then,

`Q = 8 \times 1.5`

`Q = 12 n C`

Question 8

    The surface charge density of a charged disk is 3 nC/m². If the disk has a radius of 0.5 meters, what is the charge on a circular region with a radius of 0.2 meters?

The area of a circle is

`A = \pi r^2`

Where,

A = area of the circle

r = radius of the circle

According to question 

`r = 0.2 m`

`A = \pi r^2`

`A = 3.14 \times (0.2)^2`

`A = 3.14 \times 0.04`

`A = 0.1256 m^2`

Surface charge density

`\sigma = \frac{Q}{A}`

`Q = \sigma \times A`

According to question

`\sigma = 3 \frac{n C}{m^2}`

`A = 0.1256 m^2`

`Q = \sigma \times A`

`Q = 3 \times 0.1256`

`Q = 0.3768  n C`

Question 9

    A sphere has a volume charge density of 2 nC/m³ and a radius of 0.1 meters. Find the charge enclosed within a spherical shell with a radius of 0.05 meters.

The volume of the spherical shell is 

`V = \frac{4}{3} \pi (r_2^3 - r_1^3)`

`V = \frac{4}{3} \times 3.14 \times {(0.1)^3 - (0.05)^3}`

`V = \frac{4}{3} \times 3.14 \times {0.001 - 0.000125}`

`V = \frac{4}{3} \times 3.14 \times 0.000875`

`V = \approx 0.0037 m^3`

Formula 

Charge = Volume charge density `\times` Volume

`Q = 2 \times 0.0037`

`Q = 0.0074  n C`

Question 10

    The volume charge density of a charged cylinder is 4 nC/m³, and its height is 2 meters. Calculate the charge contained in a cylindrical segment with a height of 0.8 meters.

Solution

    To calculate the charge contained in a cylindrical segment, we first need to find the volume of the segment and then multiply it by the volume charge density.

The volume of a cylindrical segment can be calculated using the formula:

`V = \pi r^2 h`

where:

V  is the volume of the segment,

 r  is the radius of the cylinder, and

 h is the height of the segment.

    Given that the height of the cylindrical segment is 0.8 meters, we need to find the radius of this segment. Since the radius of the entire cylinder isn't given, we need to use additional information or assumptions.

    If we assume that the cylinder has a uniform radius throughout, then we need to find the volume of the whole cylinder and use the given height of the segment to find the ratio of the segment's height to the cylinder's height. This ratio can be used to find the radius of the segment.

The volume of the cylinder can be calculated using the formula:

`V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}}`

Given:

`h_{\text{cylinder}} = 2` meters (height of the cylinder)

We can rearrange this formula to solve for radius r 

`r = \sqrt{\frac{V_{\text{cylinder}}}{\pi h_{\text{cylinder}}}}`

Given the volume charge density `\rho = 4 \, \text{nC/m}^3`, we can calculate the charge contained in the segment using the formula:

`Q = \rho \times V`

Now, let's perform the calculations:

1. Calculate the volume of the cylinder:

`V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}}`

2. Calculate the radius of the cylinder:

`r = \sqrt{\frac{V_{\text{cylinder}}}{\pi h_{\text{cylinder}}}}`

3. Calculate the volume of the segment:

`V_{\text{segment}} = \pi r^2 h_{\text{segment}}`

4. Calculate the charge contained in the segment:

`Q = \rho \times V_{\text{segment}}`

Let's calculate:

First, let's calculate the volume of the cylinder:

`V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}}`

Given:

`h_{\text{cylinder}} = 2 ` meters

We'll need to know the radius of the cylinder. Since it's not explicitly given, let's assume it's a uniform cylinder. If not provided, you would need additional information to determine the radius.

Now, let's calculate the radius of the cylinder:

`r = \sqrt{\frac{V_{\text{cylinder}}}{\pi h_{\text{cylinder}}}}`

To calculate the volume of the segment, we'll need the radius of the cylinder. Since we don't have this information, we'll assume a reasonable radius. Let's say r = 1 meter.

`r = 1 \, \text{meter}`

Now, we can calculate the volume of the segment using the given height of the segment:

`V_{\text{segment}} = \pi r^2 h_{\text{segment}}`

`V_{\text{segment}} = \pi (1^2) (0.8) \, \text{m}^3`

Now, calculate the charge contained in the segment using the volume charge density:

`Q = \rho \times V_{\text{segment}}`

Given:

`\rho = 4 \, \text{nC/m}^3`

`Q = (4 \times 10^{-9} \, \text{C/m}^3) \times (\pi \times 1^2 \times 0.8) \, \text{m}^3 `

Now, let's calculate this value.

`V_{\text{segment}} = \pi (1^2) (0.8) \, \text{m}^3 = 0.8 \pi \, \text{m}^3`

`Q = (4 \times 10^{-9} \, \text{C/m}^3) \times (0.8 \pi) \, \text{m}^3`

`Q = 3.2 \pi \times 10^{-9} \, \text{C}`

So, the charge contained in the cylindrical segment with a height of 0.8 meters is approximately ` 3.2 \pi \times 10^{-9}` Coulombs.