You can Download RBSE Solutions for Class 12 Physics Chapter 2 Gauss’s Law and its Applications Guide Pdf, help you to revise the complete Syllabus and score more marks in your examinations.

## Rajasthan Board RBSE Class 12 Physics Chapter 2 Gauss’s Law and its Applications

**RBSE Class 12 Physics Chapter 2 Text Book Exercise with Answers**

### RBSE Class 12 Physics Chapter 2 Multiple Choice Type Questions

Question 1.

Electric field intensity is maximum due to a uniformly charged solid non-conducting sphere :

(a) at centre

(b) between a point on the surface and the centre

(c) at surface

(d) at infinity

Answer:

(c) at surface

At surface, electric field intensity is given by

Question 2.

Energy density (in vacuum) at a place where electric field intensity E is :

Answer:

(d)

Energy density (u_{d})

u_{d} = \(\frac{1}{2}\) ε_{0}E^{2}

Question 3.

A charge of 1 pCis placed at the centre of a cube of side 0.2 m. The electric flux leaving each vertex of the cube in V/m is :

(a) 1.12 × 10^{4}

(b) 2.2 × 10^{4}

(c) 1.88 × 10^{4}

(d) 3.14 × ^{4}

Answer:

(c) 1.88 × 10^{4}

Leaving out the flux from each surface of cube

\(\frac{1}{6} \frac{q}{\varepsilon_{0}}\)

Question 4.

Two dipoles of charges ±q are placed normally inside a cube. Then the total electric flux leaving out the cube will be :

Answer:

(c)

Due to dipole, there is no area from which flux is passing. Thus the leaving out flux = 0.

Question 5.

On giving a negative charge to a soap bubble, its radius :

(a) decreases

(b) increases

(c) remain unchanged

(d) data inadequate

Answer:

(b) increases

On giving negative charge, due to increase in surface area, radius increases.

Question 6.

A charge q is in a sphere and the electric flux leaving out is \(\frac{q}{\varepsilon_{0}}\) How much would be the change in the electric flux on reducing the radius to its half?

(a) Becomes 4 times the initial value

(b) Becomes \(\frac{1}{4}\) the initial value

(c) Becomes half of the initial value

(d) Remains unchanged

Answer:

(d) Remains unchanged

Leaving out electric flux does not depend on radius.

Question 7.

The value of electric flux leaving out from a unit positive charge in vacuum is :

(a) ε_{0}

(b) ε_{0}^{-1}

(c) (4πε_{0})^{-1}

(d) 4πε_{0}

Answer:

(b) ε_{0}^{-1}

Total emerging out the flux from unit charge

Question 8.

The radii of two conducting sphere are a and b. They are charged by equal amount then the ratio of electric field intensity at its surfaces :

(a) b^{2} : a^{2}

(b) 1 : 1

(c) a^{2} : b^{2}

(d) b : a

Answer:

(a) b^{2} : a^{2}

Question 9.

The radii of two conducting spheres are a and b. They are charged by equal charge density. What would be the ratio of the electric field intensities at their surface?

(a) b^{2} : a^{2}

(b) 1 : 1

(c) a^{2} : b^{2}

(d) b : a

Answer:

(b) 1 : 1

Question 10.

The graph between a electric field intensity and its variation with \(\frac{1}{r}\) due to a long straight charged wire is:

Answer:

(c)

Thus graph (c) will be correct

Question 11.

A square is placed in a uniform electric field E parallel to the horizontal such that the plane of the square makes an angle 30° with the field. If the side of the square is a, then the flux through the square wire be :

(a) \(\frac{\sqrt{3} E a^{2}}{2}\)

(b) \(\frac{E a^{2}}{2}\)

(c) zero

(d) none of these

Answer:

(c) zero

Since electric field is acting along, plane of the square, so the angle between area vector \(\vec{A}\) and \(\vec{E}\) is 90°. These give ϕ = EAcosθ = EAcos90 = 0.

### RBSE Class 12 Physics Chapter 2 Very Short Answer Type Questions

Question 1.

When is the electric flux leaving out the elemental area placed in electric field \(\vec{E}\) zero?

Answer:

If the area of elemental surface \(\overrightarrow{d s}\) is at 90° to electric field, then emergent out flux will be zero.

Question 2.

At what conditions, is the electric field intensity due to a uniformly charged non-conducting sphere zero?

Answer:

Uniform charged insulating sphere has electric field intensity zero at centre and infinity.

Question 3.

Write the formula for force acting on a unit area of a charged conductor. Also give its direction.

Answer:

Force, F = \(\frac{\sigma^{2}}{2 \varepsilon_{0}}\)

The direction of force will be perpendicular to surface, outwards.

Question 4.

Where is the energy stored due to an electric charge?

Answer:

Energy is stored in the volume of electric field.

Question 5.

A conducting sphere of diameter d is charged with Q. What is the value of electric field inside the sphere?

Answer:

As whole charge comes on the surface so inside electric field will be zero.

Question 6.

If in Coulomb’s law, there would be dependence on \(\frac{1}{r^{3}}\) in place of \(\frac{1}{r^{2}}\), then would Gauss’s law be true?

Answer:

No, Gauss’s law will be applicable for those fields for which they follow inverse square law.

Question 7.

If the net charge enclosed in a Gaussian surface is positive, then what is the nature of total electric flux through the surface?

Answer:

The leaving out flux will be positive.

Question 8.

If the electric flux leaving out the closed surface is zero, then what could be said about the surface?

Answer:

- Total charge on the surface is zero.
- ϕ
_{emergent}= ϕ_{entering}

Question 9.

If the net charge inside a Gaussian surface is zero, then does it mean that electric field intensity at each point on the surface is zero?

Answer:

No, because

Question 10.

Define linear charge density?

Answer:

Charge per unit length is known as linear charge density (λ).

λ = \(\frac{q}{l}\)

Question 11.

What would be the change in electric field for a charged sheet of (a) surface charge density to move from one side to another?

Answer:

Change in electric field

= E_{1} – E_{2}

\(=\frac{\sigma}{2 \varepsilon_{0}}-\left(-\frac{\sigma}{2 \varepsilon_{0}}\right)=\frac{\sigma}{\varepsilon_{0}}\)

Question 12.

Plot a graph between electric field intensity due to a uniformly charged non-conducting sheet and distance.

Answer:

The electric field is independent to distance.

Question 13.

What is the electric field intensity at the centre of the uniformly charged non-conducting sphere?

Answer:

As charge inside insulator sphere is zero, therefore intensity of electric field will be zero.

Question 14.

If charge q is situated at the centre of a sphere. Now, if the charge is placed inside the cylindrical surface of same volume. Then what would be the ratio of flux leaving out the surface in both the conditions?

Answer:

In both conditions, the ratio of electric flux will be equal.

### RBSE Class 12 Physics Chapter 2 Short Answer Type Questions

Question 1.

Explain electric flux. Write its SI unit and dimensions.

Answer:

Electric Flux

Area of a plane surface is a vector quantity; because it has magnitude and direction both. The direction of area vector is in the direction of normal to the surface. In the adjoining figure 2.1, a small area ds is shown in vector form i. e.,

\(\overrightarrow{d s}=|\overrightarrow{d s}| \hat{n}=d s \hat{n}\)

Where ds is the magnitude of area vector and \(\hat{n}\) is unit vector along the normal on the surface of area.

Electric Flux: “Number of electric field lines of force passing normally through the surface placed in the electric field is known as the electric flux linked with the surface.” It is denoted by ϕ_{E} and it is a scalar quantity.

If a surface of area S is placed in uniform electric field of intensity E such that its plane remains at right angle to field lines [fig. 2.2 (a)], the electric flux linked with the surface,

If the surface is placed such that the normal on the surface that is area vector \(\overrightarrow{\mathrm{S}}\) makes an angle θ with the field lines then flux attached with the surface,

ϕ_{E} = ES cosθ

or ϕ_{E} = \(\vec{E} \vec{S}\) ………..(2)

“Thus the dot product of electric field vector and area vector provides the electric flux linked with the surface.”

If the surface is placed in uniform electric field then the whole area of the surface is divided in many small elements of area and by adding the electric fluxes linked with these elements we can obtain the total flux i.e.,

The integral used in above equation is known as ‘Closed surface integral.’ This integration tells that area S is divided in many small areal elements \(\overrightarrow{d s}\) and electric flux linked with each element is \(\vec{E} . \overrightarrow{d s}\). By adding all these flux elements, the total flux attached with whole surface is obtained which is given by equation (3).

Again from equation (2) it is clear that electric flux is a scalar quantity because it is \(\vec{B}\) dot product of \(\vec{E}\) and \(\overrightarrow{d s}\). The flux linked with small element of area may be positive, zero or negative, which depends upon the angle θ between \(\vec{E}\) and \(\vec{S}\). If θ < 90°, then flux will be positive, for θ = 90° flux will be zero and for θ > 90°, the flux will be negative.

Question 2.

Explain linear charge density. Write it’s unit.

Answer:

Linear charge density : When the charge density is along a line, then the quantity of charge per unit length is called linear charge density and it is represented by λ.

λ = \(\frac{q}{l}\), where q is charge uniformly distributed on length l.

Unit of λ = Cm^{-1}

Example : If q charge is uniformly distributed on a ring of radius R, then the linear charge density of ring will be \(\frac{q}{2 \pi R}\)

Question 3.

Explain surface charge density. Write it’s unit.

Answer:

Surface charge density : When the charge is distributed on a plane or curved surface, then charge per unit area is called surface charge density of charge and represented by σ.

∴ σ = \(\frac{q}{A}\)

Where q is charge on surface of area A

∴ Unit of σ = Cm^{-2}

Example : If q charge is distributed uniformly on sphere of radius R, then surface charge density of the sphere \(\sigma=\frac{q}{4 \pi R^{2}}\).

Question 4.

Explain volume charge density. Write its unit.

Answer:

Volume charge density : When the charge density is throughout the volume of a body, then the charge per unit volume is called volume charge density and it is represented by ρ.

∴ ρ = \(\frac{q}{V}\)

Hence the unit of ρ is Cm^{-3}

Example : If a charge q is uniformly distributed in the whole volume of a sphere of radius R, then \(\rho =\frac { q }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } =\frac { 3q }{ 4\pi { R }^{ 3 } } \)

Question 5.

Explain Gauss’s law for static electricity.

Answer:

Gauss’s Theorem or Gauss’s Law

Gauss’s law is another form of Coulomb’s law relating electrostatic forces. This theorem represents the relation between electric flux passing normally through an imaginary closed and self decided surface and the charge enclosed by the surface. This imaginary and self decided closed surface is called ‘Gaussian surface.’ With the help of this theorem, electric field due to any charge or group of charges can be determined easily.

Condition : It is applicable for closed charges.

Statement : According to Gauss’s theorem, “the electric flux passing through a closed surface is \(\frac{1}{\varepsilon_{0}}\) times the charge bounded by the surface.”

Mathematically it can be represented as,

\(\phi_{E}=\oint_{s} \vec{E} \cdot \vec{d} s=\frac{1}{\varepsilon_{0}} \Sigma q\) ………………… (1)

Where Σ_{q} is total, bound charge and ε_{0} is permittivity of free space.

If total of bounded charge is zero or there is no charge inside the surface, then

\(\phi_{E}=\oint_{S} \vec{E} \cdot \overrightarrow{d s}=0\) …………….. (2)

Proof : (1) When the charge is situated inside the surface : Suppose there is a closed surface of area S and a charge +q is situated at point O inside the

surface (fig. 2.7). The complete surface can be supposed to be made up of infinitely small elements. At point P on the surface, the area of considered element is \(\overrightarrow{d s}\) and \(\vec{E}\) is the intensity of electric field 0 is

the angle between \(\vec{E}\) and \(\overrightarrow{d s}\). The distance of point P from charge + q is r.

∴ E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}\)

Now the electric flux passing through the element

If there are q_{1},q_{2},………, q_{n} charges enclosed by the surface and ϕ_{1}, ϕ_{2}, …….. ϕ_{n} are the values of flux relating to each charge, then

(2) When charge is placed outside the surface : Suppose charge +q is situated outside the surface at point O. A cone of solid angle dm is drawn such that it intersects the closed surface at four places situated around points P, Q, R and S as shown is fig. 2.8. The electric flux linked with these places are dϕ_{1}, dϕ_{2}, dϕ_{3} and dϕ_{4} respectively.

Question 6.

Why is charge always on the outer surface of the conductor?

Answer:

As electric field inside charged sphere is zero (E = 0), therefore emergent flux from Gaussian surface will be zero.

or ϕ = 0 but, ϕ= \(\frac { q }{ { { \varepsilon }_{ \o } } } \) ⇒ q = ϕε_{0} = 0

It means, charge inside surface will be zero. Thus, the whole charge Q will be distributed on its surface.

Question 7.

Why does the size of a soap bubble increases on charging it?

Answer:

There is equilibrium inside positive and negative charges of soap bubble. When it is given negative charge, then the position of equilibrium start to change. To achieve this equilibrium, charges try to repel each other. Thus the surface area of bubble increases. Due to increase in surface area, the bubble increases in size.

Question 8.

Obtain the formula for electrostatic force and electric pressure on the surface of the charged conductor.

Answer.

Force on the Surface of Charged Conductor

The charge provided to a conductor is uniformly distributed on its surface. A repulsive force acts by the charge on the rest part from the charge present at small part of the conductor and this way, a force of repulsion acts at each small element on the conductor, and the total force on the surface of the conductor is the vector sum of force acting on all the small elements.

That is why the charged conductor experiences pressure outwards the charged conducting surface.

Let the surface charge density on the surface of the conductor be σ. We will consider points outside and inside the conductor, two identical points P_{1} and P_{2} respectively (see fig. 2.20).

Since, the electric field outside the charged conducting surface is \(\frac{\sigma}{\varepsilon_{0}}\). Thus, electric field at point P_{1}

\(E_{P_{1}}=\frac{\sigma}{\varepsilon_{0}}\) …………….. (1)

Electric field inside the conductor is zero. Thus, electric field at P_{2}

Ep_{2} = 0 …………..(2)

Now, we will divide the conductor into two parts :

(i) Element AB whose surface is ds and

(ii) The remaining part ACB

If the electric field intensity at near point due to element ABis \(\vec{E}_{1}\) and \(\vec{E}_{2}\) due to ACB part. Then, from the fig. 2.20,

Ep_{1} = E_{1} + E_{2} …………… (3)

(E_{1} and E_{2} are in same direction at P_{1})

and Ep_{2} = E_{1} – E_{2} …………… (4)

(E_{1} and E_{2} are in opposite direction at P_{2})

From eqns. (2) and (4)

E_{1} – E_{2} = 0

i.e., E_{1} = E_{2} …………………. (5)

From eqns. (1), (3) and (5)

Question 9.

Obtain the relation for stored energy in unit volume of electric field.

Answer:

Energy per Unit Volume in an Electric Field

Electrostatic force is directed normally outwards on the surface of the charged conductor. To increase the amount of charge on conductor or to increase the volume of electric field, an opposite work against force is done that is stored as energy in the electric field.

For simplicity, we consider a spherical charge of radius r. The surface charge density on sphere is σ as shown in figure 2.21.

Pressure outwards on the surface of the sphere

P = \(\frac{\sigma^{2}}{2 \varepsilon_{0}}\) ……….. (1)

Thus, force outwards on the spherical surface

Work done to press the sphere against the force by the distance dr,

Decrease in volume of sphere due to compression (or increase in volume of electric field)

Energy stored in whole system due to electric field

And energy stored in per unit volume of electric field or energy density

If there is any other medium in place of vacuum or air, then

The above formulae are obtained by taking the example of spherical shell but these are widely valid.

Question 10.

Obtain the formula of maximum surface charge density for equilibrium of charged soap bubble.

Answer.

Equilibrium Condition for Charged Soap Bubble

The pressure of air on the internal surface of the soap bubble is more than that the atmospheric pressure present on the external surface. This pressure is balanced by the force of surface tension. If the radius of the bubble is r and surface tension is T, then the excess pressure is

P_{ex} = \(\frac{4 T}{r}\) ………….. (1)

Now, if the bubble is charged by the surface charge density of σ, then the electric pressure of \(\frac{\sigma^{2}}{2 \varepsilon_{0}}\) acts outward on the surface. In this case,

Thus a state will come when excess pressure will become zero on charging the bubble. After it, the bubble blasts. Thus for equilibrium,

P_{ex} = \(\frac{4 T}{r}-\frac{\sigma^{2}}{2 \varepsilon_{0}}\) = 0

Question 11.

Verify Gauss’s law from Coulomb’s law.

Answer:

Derivation of Gauss’s Theorem by Using Coulomb’s Law :

According to definition of electric flux,

\(\phi_{E}=\oint_{S} \vec{E} \overrightarrow{d s}=\oint_{S} E d s \cos \theta\)

The intensity of electric field \(|\vec{E}|\) at same distance from charge q remains constant and for spherical surface θ = 0°.

According to Coulomb’s law. intensity of electric field at distance r from charge q.

This is Gauss’s theorem.

Question 12.

You are going in a car. It is about to lightning, then how will you protect yourself?

Answer:

We will close the window of the car. So that entire charge will lie on the surface of car.

Question 13.

The linear charge densities on two straight parallel long linear charges are λ_{1} and λ_{2}. Obtain the formula for force acting per unit length between them.

Answer:

Electric field intensity at distance from a wire of linear charge density λ_{1}

As second conductor is also in this field. Therefore force acting on its unit length (Charge q = λ_{2})

Question 14.

The charge densities on planar parallel bases of two infinite expansions are +σ and -σ respectively. What would be the electric field intensities at a point between them?

Answer:

On point between both plates, the electric field produced due to both plates \(\vec{E}_{1}\) and \(\vec{E}_{2}\) will be in same direction.

∴ Resultant electric field

### RBSE Class 12 Physics Chapter 2 Long Answer Type Questions

Question 1.

Calculate electric field intensities on charging a spherical conductor of radius R with charge q in following conditions :

(a) r > R

(b) r < R

(c) on the spherical surface

(d) at the centre of sphere

Plot a graph between electric field intensity and distance.

Answer:

Electric Field Intensity due to a Uniformly Charged Conducting Sphere

The charge given to a conductor always rests on its surface. Therefore charge given to a spherical conductor will rest on its surface; Hence the charged spherical conductor will behave as spherical shell. Therefore the expressions for electric field will same as those for shperical shell, i.e.,

(i) Intensity of electric field outside the sphere at distance r from its centre;

Question 2.

Calculate electric field intensity due to uniformly charged non-conducting sphere :

(a) outside the sphere

(b) at the surface of sphere

(c) inside the sphere

(d) at the centre of the sphere

Plot at graph between electric Held intensity and distance.

Answer:

Electric Field Intensity due to a Uniformly Charged Non-conducting Sphere

When charge is given to non-conducting sphere, it uniformly spreads throughout its volume. If q is the charge given and R is the radius of the sphere, then the volume charge density

(a) Outside the sphere : In this case taking O as centre and r as radius, a spherical Gaussian surface is drawn. The point P will be situated at this surface.

The direction of \(\vec{E}\) will be outwards directed due to symmetry. The charge enclosed by the Gaussian surface

(b) When point P is situated on the surface of the sphere : In this case we can consider the surface of the sphere as Gaussian surface. Therefore the whole charge of the sphere will again be charge enclosed. Thus by putting r = R in equation (5), we get

(c) When point P is inside the sphere (r < R):

Here also we draw a spherical Gaussian surface taking O as centre and r as radius. Point P will be situated at this surface. Now the charge enclosed by this Gaussian surface,

Question 3.

Find the electric field intensity due to infinite a uniformly charged wire at a point near it using Gauss’s law. Plot the curve between electric field intensity and distance.

Answer:

Electric Field Intensity due to an Infinite Long Line Charge from Gauss’s Law

Let the linear charge density is λ and the intensity of electric field at point P is to be determined.

To apply Gauss’s law, we have to decide the direction of electric field at P. For this purpose let us consider two small elements of length dλ. A and B at equal distance from O as shown in the figure 2.9. The intensities of electric field due to these elements at P, dE_{1} and dE_{2} are equal in magnitude. If the electric fields are resolved in two normal components, then the components along OP provide the electric field \(\vec{E}\) along OP and normal components dE_{1} sinθ and dE_{2} sinθ cancel out each other. Similar will be the result for other pairs of length elements considered.

Thus the direction of electric field at P will be in OP direction i.e., normal to linear charge.

Calculation of Electric Field : For application of Gauss’s law we require a Gaussian surface. For this purpose, we draw a cylindrical Gaussian surface of radius r and axis coinciding the linear charge AB. Point P lies on the curved surface of the cylinder. The enclosed charge

Σq = λl

where l is the length of the cylinder. Therefore from Gauss’s law,

Where \(\hat{r}\) unit vector in direction OP i.e., in perpendicular to linear charge.

From equation (3)

E ∝ \(\frac{1}{r}\)

Therefore graph between E and r will E be as shown in adjoining figure 2.11

Question 4.

Find the electric field intensity due to infinite uniformly charged non-conducting sheet at a point near it, using Gauss’s law. Explain the dependence of electric field intensity.

Answer:

Electric Field due to an Infinite Uniformly Charged Non-conducting Sheet

Direction of electric field due to infinite charged sheet : Suppose σ is the surface charge density on the charge sheet and at point P we have to find the intensity of electric field. To apply Gauss’s theorem we require the direction of electric field at P. For this purpose we consider two small surface elements S_{1} and S_{2} the same distance from O as shown in the figure 2.12. The components dE_{1} cosθ and dE_{2} 2cosθ add to provide resultant field while perpendicular components dE_{1} sinθ and dE_{2}sinθ being equal and opposite direction so they are cancel out each other. Same will the case for other considered pairs of elements. Thus it is concluded that the electric field is in perpendicular direction and away from the sheet.

Formula for Electric Field :

To determine the electric field at point P, we draw a Gaussian surface. According to figure we draw a cylindrical Gaussian surface of area of cross-section S such that its one place surface S_{1} passes through P

and the plane. Surface S_{2} is on the other side of the charge sheet.

The enclosed charge,

Σq = σS where σ is surface charge density

Where \(\hat{n}\) is unit vector in OP direction thus,

(i) E does not depend on r.

(ii) If the sheet is positively charged (σ > 0), then electric field will be directed away.

(iii) If the sheet is negatively charged (σ < 0), then electric field will be directed towards. Dependence of electric field :

(a) The value of E does not depends up on distance (r).

(b) ∵ E = \(\frac{\sigma}{2 \varepsilon_{0}}\) ⇒ E ∝ σ

Both the these dependencies are shown in following graphs (fig. 2.14) :

Question 5.

Determine the direction of electric field intensity at a point near a uniformly charged infinite conducting plate. Obtain the relation for electric field intensity using Gauss’s law. Draw the required diagram.

Answer:

Electric Field Intensity due to a Uniformly Charged Infinite Conducting Plate

The basic difference between the non-conducting sheet and conducting sheet is that the charge in a non-conducting sheet remain at a place where the charge is given while the charge in a conducting sheet is spread out in a whole sheet and the electric field inside the conducting sheet is zero.

Value of electric field : In the figure 2.15(a) conducting sheet is shown. When the some charge is given to the conducting plate, it get’s distributed on the entire external surface of the conductor. If the plane charge sheet is of uniform thickness and infinite size, then the surface charge density (σ) of the charge being uniform on the entire surface. It is the same at the both surfaces.

Consider a cylindrical surface perpendicular to the sheet with the area of cross-section (s) as shown in figure 2.15 (a). On the circular surface of the cylinder P is located.

The total charge enclosed = Σq = σS

By the Gauss’s law

Where \(\hat{n}\) is the unit vector perpendicular to the sheet from the equation (3). We see that the electric field at any point due to the sheet does not depend upon the distance of the point from the sheet i.er the magnitude of electric field at all the points is equal.

∴ E ∝ r^{0} ………… (4)

The electric field intensity in any other medium.

∴ E = \(\frac{\sigma}{\varepsilon_{m}}=\frac{\varepsilon}{\varepsilon_{r} \varepsilon_{0}}=\frac{\sigma}{K \varepsilon_{0}}\) ………… (5)

Graph of electric field : The electric field intensity does not depend upon the distance of point from the uniformly charged sheet i.e, it remains thhe same at all the points. The graph of it is obtained as shown in the figure 2.15 (b).

### RBSE Class 12 Physics Chapter 2 Numerical Questions

Question 1.

In a closed surface, the electric flux entering and leaving out the surface are 400 Nm^{2}/C and 800 Nm^{2}/C respectively. What would be the charge enclosed by the surface?

Answer.

The electric flux entering in surface

ϕ_{1} = -400Nm^{2} / C

Electric flux leaving out through surface

ϕ_{2} = 800Nm^{2}/C

∴ Total flux linked with the closed surface

ϕ = ϕ_{1} + ϕ_{2} = -400 + 800 = 400Nm^{2} / C

From Gauss’s theorem,

ϕ_{E} = \(\frac{1}{\varepsilon_{0}}\) Σq

Σq = ϕ_{E} × ε_{0}

= 400 × 8.86 × 10^{-12}

= 3.54 × 10^{-9} C

= 3.54 nC

Question 2.

A uniformly charged sphere of diameter 2.4 m has surface charge density of 80 pC/m^{2}. Determine the total electric flux leaving out the spherical surface along with the spherical charge.

Answer:

(a) Surface charge density on the surface of uniformly charged sphere

σ = \(\frac{q}{A}=\frac{q}{4 \pi r^{2}}\)

q = 4πr^{2}σ

= 4 × 3.14 × 1.2 × 1.2 × 80.0 × 10^{-6} C

= 1447 × 10^{-6} C

= 1.447 × 10^{-3} C

= 1.45 × 10^{-3} C = 1.45 mC

b) From Gauss’s law if we consider surface of sphere as Gausssian surface then total charge on sphere will be bound to sphere.

∴ Leaving out flux through surface

Question 3.

Calculate the total electric flux and flux related to each vertices for a cube of side a when charge q is placed at

(i) centre of cube

(ii) an end of the cube

(iii) a plane of cube.

Answer:

(i) As in cube, there are 6 faces of equal

area.

∴ The leaving out flux through each face of cube

As each face of cube is in symmetry.

(ii) As in cube there are four corners on each surface. Therefore total flux bound through cube = \(\frac{q}{4 \varepsilon_{0}}\)

∴ The leaving out flux through any one face

(iii) As in cube each surface has two part

∴ Flux bound to cube = \(\frac{q}{2 \varepsilon_{0}}\)

∴ The leaving out flux through any one face of the cube

Question 4.

The electric field intensity at a distance of 20 cm from the centre of a sphere is 10 V/m. The radius of the sphere is 5 cm. Determine the electric field intensity at a point 8 cm distance from the centre of the sphere.

Answer:

Electric field intensity on any external point of sphere

Question 5.

An infinite linear charge at 2 cm distance produces an electric field of 9 × 10^{4} N/C. Determine the linear charge density.

Answer:

Electric field intensity due to linear charge of infinite extension

E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \lambda}{r}\)

Where, λ is linear charge density and r is distance of observation point from charge. According to question,

Question 6.

In a figure, a charge of +10 μC is placed just 5 cm above the centre of a square of side 10 cm. What would be the electric flux passing through the square?

Answer:

We consider a cubical Gaussian surface of side 10 cm. Its centre lie where charge q = 10 μC placed.

∴ According to Gauss’s theorem, the leaving out flux through Gaussian surface

ϕ_{E} = \(\frac{q}{\varepsilon_{0}}\)

According to Gauss’s theorem, the leaving out flux through Gaussian surface

∴ The leaving out flux through each face

Question 7.

The area of a metallic plate is 10^{-2} m^{2} .A charge of 10 μC is given to the plate. Determine the electric field intensity at a point near to the plate.

Answer:

Given A = 10^{-2} m^{2}

q = 10 μC = 10 × 10^{-6} C

Electric field intensity near plate

Question 8.

Two metallic surface of area 1 m are placed at 0.05 m parallel to each other. Both have same magnitude of charges but opposite in direction. If the electric field between them is 55 V/m, then what would be charge on each?

Answer:

At the point C between both the plates, the electric field intensity due to both plates \(\vec{E}_{1}\) and \(\vec{E}_{2}\) will be in same direction.

Therefore, resultant electric field

Question 9.

A particle of mass 9 × 10^{-5}g is placed on a uniformly charged long horizontal sheet of surface charge density of 5 × 10^{-5}C/m^{2}. How much should be the charge on the particle so that it would not fall when left freely?

Answer:

Given, m = 9 × 10^{-5}g = 9 × 10^{-8}kg

σ = 5 × 10^{-5}C/m^{-2}

As, qE = mg

Question 10.

In X-Y plane, there is surface charge density of 5 × 10^{-16} C/m^{2} on a long uniformly charged sheet. A circular loop of radius 0.1 m makes an angle of 60° with Z-axis. Determine the electric flux through the loop.

Answer:

Given, σ = 5.0 × 10^{-16} C/m^{-2}, r = 0.1 m,

θ = 60° electric field due to layer of plane charge

E = \(\frac{\sigma}{2 \varepsilon_{0}}\)

Flux passing through circular surface

Question 11.

An electron of energy 10^{3} eV is fired normally to the infinite and conducting plate from 5 mm. Calculate the minimum surface charge density on the conducting plate so that the electron does not strike the plate.

Answer:

Given, K = 10^{2}eV, d = 5mm = 5 × 10^{-3} m

σ = ?

∵ Kinetic energy, K = eV

⇒ 10^{3}eV = eV

∴ V = 10^{3} Volt

Question 12.

The pressure inside and outside the soap bubble is equal. The surface tension on the soap bubble is 0.04 N/m and the diameter of the bubble is 4 cm. Determine the charge on the soap bubble.

Answer:

Given, T = 0.04 N/m

Diameter = 4 cm = 4 × 10^{-2} m

∴ Radiu r = 2 cm = 2 × 10^{-2} m

Charge q = ?

## Leave a Reply