Electric Fields and Gauss' Law MCQ Quiz - Objective Question with Answer for Electric Fields and Gauss' Law - Download Free PDF

Explanation:

For a point outside sphere, we have,

\(E=\frac{Q}{4 \pi \varepsilon _{o} R^{2}}\)

\(1. Q \rightarrow Q, R \rightarrow R \Rightarrow E_{1}=E\)

\(2. Q \rightarrow 2Q, R \rightarrow R \Rightarrow E_{2}=2E\)

\(3. Q \rightarrow Q', R \rightarrow R \Rightarrow E_{3}=\rho R/3\epsilon_0 \\ \Rightarrow E_3=\frac{4Q}{4/3\pi (2R)^3}\frac{R}{3\epsilon_0} =E/2\)

\(\Rightarrow E_{2}>E_{1}>E_{3}\)

Concept:

• The electric field is defined as the force per unit of positive charge.
• The SI unit of electric field intensity is N/C.
• The electric field intensity due to infinite long thin straight wires is given by, \(E=\frac{λ}{2\pi \epsilon_0 R}\)
• The electric field intensity due to infinite long thin straight wire is inversely proportional to the distance at which the electric field is calculated.
• Electric field intensity, E∝ \(\frac{1}{R}\)

Explanation: Draw a diagram according to the question detail,

The electric field intensity at the midpoint due to the infinite wire of uniform linear charge density 2λ  is,

\(E_1=\frac{2λ}{2\pi \epsilon_0 (R/2)}=\frac{4λ}{2\pi \epsilon_0 R}=\frac{2λ}{\pi \epsilon_0 R}\)

The electric field intensity at the midpoint due to the infinite wire of uniform linear charge density 3λ is

\(E_2=\frac{3λ}{2\pi \epsilon_0 (R/2)}=\frac{6λ}{2\pi \epsilon_0 R}=\frac{3λ}{\pi \epsilon_0 R}\)

The net electric field intensity at the midpoint is, 

\(E_n=E_2-E_1\)

\(E_n=\frac{3λ}{\pi \epsilon_0 R}-\frac{2λ}{\pi \epsilon_0 R}=\frac{λ}{\pi \epsilon_0 R}\)

Concept:

• When there are various types of charges in a region, but the total charge is zero, the region is supposed to contain a number of dipoles. 
• Therefore, at points outside the region, the dominant electric field is inversely proportional to the cube of r for a large value of r 
• The electric field due to the dipole is 

\(E = \frac{q}{4 \pi \epsilon_0}\frac{2P}{r^3}\)

Where p = q × l (dipole)

r is the distance from the dipole r >> l

Hence the correct option is option 3

Calculation:

Let the charge +q be placed at the origin (x = 0) and the charge +9q be placed at x = d.

Let the electric field vanish at a point x (0 < x < d) on the x-axis, between the two charges.

Electric field due to +q at point x:

E_q = kq / x²

Electric field due to +9q at point x (distance from +9q is d - x):

E_9q = k(9q) / (d - x)²

For the net electric field to vanish:

E_q = E_9q

⇒ kq / x² = k(9q) / (d - x)²

Canceling k and q from both sides:

1 / x² = 9 / (d - x)²

Taking square roots:

1 / x = 3 / (d - x)

Cross-multiplying:

d - x = 3x

⇒ d = 4x

⇒ x = d / 4

Explanation:

An electric dipole in a uniform electric field experiences a torque that tends to align the dipole with the electric field. The torque ( $\overrightarrow{\tau}$ ) acting on an electric dipole moment ( $\overrightarrow{P}$ ) placed in a uniform electric field ( $\overrightarrow{E}$ ) is given by the vector cross product of the dipole moment and the electric field.

The formula for the torque is:

$\overrightarrow{\tau} = \overrightarrow{P} \times \overrightarrow{E}$

The cross product of two vectors results in a vector that is perpendicular to both, and its magnitude is given by the product of the magnitudes of the two vectors and the sine of the angle between them.

Therefore, the correct answer is option 1: $\overrightarrow{P} \times \overrightarrow{E}$ .

∴ The torque acting on the electric dipole is $\overrightarrow{P} \times \overrightarrow{E}$ .

CONCEPT:

Electric Field Intensity:

• The electric field intensity at any point is the strength of the electric field at the point.
• It is defined as the force experienced by the unit positive charge placed at that point.

\(\vec E = \frac{{\vec F}}{{{q_o}}}\)

Where F = force and qo = small test charge

• The magnitude of the electric field is

\(E = \frac{{kq}}{{{r^2}}}= \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}\)

Where K = constant called electrostatic force constant, q = source charge and r = distance

• The region around a charged particle in which electrostatic force can be experienced by other charges is called the electric field.
  • The electric field is denoted by E.

EXPLANATION:

• Electric field intensity is a vector quantity because it can only be properly defined when its magnitude and direction both are known. So option 2 is correct.

CONCEPT:

• Gauss's Law: Gauss's law for the electric field describes the static electric field generated by a distribution of electric charges. 
  • It states that the electric flux through any closed surface is proportional to the total electric charge enclosed by this surface. 

\(\phi_E = \frac{Q}{\epsilon_o}\)

Where ϕ_E = electric flux through a closed surface S enclosing any volume V, Q = total charge enclosed with V, and ϵ_o = electric constant

• Electric flux is the flow of the electric field through a given area.

• Electric flux is proportional to the number of electric field lines going through a virtual surface.

\({\phi}_E= E\cdot S = EScos\theta\)

Where E = electric field, S = are of the surface, E = magnitude, θ = angle between the electric field lines and the normal (perpendicular) to S, and ϕ_E = flux the electric field through a closed cylindrical surface.

EXPLANATION:

• According to Gauss's law

 \(ϕ_E = \frac{Q_(enclosed )}{ϵ_o}\)

• If the radius of the Gaussian surface is doubled, the outward electric flux will remain the same.
• This is because electric flux depends only on the charge enclosed by the surface.

​option 4 is correct.

CONCEPT:

Gauss's law:

• According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is \(\frac{1}{ϵ_o}\) the charge enclosed by the closed surface.

\(\Rightarrow ϕ=\frac{Q}{ϵ_o}\)

Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵo = permittivity

Important points:

1. Gauss’s law is true for any closed surface, no matter what its shape or size.
2. The charges may be located anywhere inside the surface.

EXPLANATION:

Gauss's law:

• According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is \(\frac{1}{ϵ_o}\) the charge enclosed by the closed surface.
• So if the total charge enclosed in a closed surface is Q, then the total electric flux associated with it will be given as,

\(\Rightarrow ϕ=\frac{Q}{ϵ_o}\)     -----(1)

• By equation 1 it is clear that the total flux linked with the closed surface in which a certain amount of charge is placed does not depend on the shape and size of the surface. Hence, option 4 is correct.

The correct answer is option 1) i.e. Φ/2

CONCEPT:

Gauss's Law for electric field: It states that the total electric flux emerging out of a closed surface is directly proportional to the charge enclosed by this closed surface. It is expressed as:

\(ϕ = \frac{q}{ϵ_0}\)

Where ϕ is the electric flux, q is the charge enclosed in the closed surface and ϵ₀ is the electric constant.

EXPLANATION:

Given that:

Consider a charge 'q' placed inside a cube of side 'a'.

The electric flux according to Gauss's law, 

\(\Rightarrow ϕ = \frac{q}{ϵ_0}\)

If the charge enclosed is halved, then  

\(\Rightarrow q' =\frac{q}{2}\)

Therefore, the new electric flux associated with this, 

\(\Rightarrow ϕ' =\frac{q'}{\epsilon_0}= \frac{\frac{q}{2}}{ϵ_0} = \frac{1}{2}\times \frac{q}{\epsilon_0} \)

\(\Rightarrow \phi '= \frac{1}{2} \times \phi = \frac{\phi}{2}\)

The electric flux emerging from a closed surface is independent of the shape or dimensions of the closed surface.

Concept:

Electric flux (ϕ): The number of electric field lines passing through a surface area normally is called electric flux. It is denoted by Φ.

The electric flux through a chosen surface is given by:

\({\rm{\Delta }}ϕ = \vec E.{\rm{\Delta }}\vec S = E{\rm{\Delta }}Scos\theta \)

Where θ is the angle between the electrical field and the positive normal to the surface.

Gauss’s Law: It states that the net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ0.

\(\oint \vec E.\overrightarrow {ds} = \frac{{{q_{inside}}}}{{{ϵ_0}}}\)

Where E = electric field, ds = small area, qinside = the total charge inside the surface, and ϵ0 = the permittivity of free space.

Calculation:

According to the question, the charge is placed on one of the corners of a cube, and to cover up a charged particle fully we need 8 cubes.

\(\therefore ϕ =\frac{1}{8} \frac{q}{{{\varepsilon _0}}}\)

Therefore the electric flux passing through the given cube is  

 \(\Rightarrow ϕ = \frac{q}{{8{\varepsilon _0}}}\) 

CONCEPT:

Gauss law:

• According to this law, the total flux linked with a closed surface called Gaussian surface is 1/ϵ_o times the charge enclosed by the closed surface.

\(\Rightarrow \phi=\oint \vec{E}.\vec{dA}=\frac{Q}{\epsilon_o}\)

• This law is applicable to all the closed surfaces.

EXPLANATION:

• According to this law, the total flux linked with a closed surface called Gaussian surface is 1/ϵo times the charge enclosed by the closed surface.
• So, this law is applicable to all the closed surfaces. Hence, option 1 is correct.

CONCEPT: 

• Dipole: The measure of the polarity of the system is the electric dipole moment.
• The simplest example of an electric dipole is a pair of electric charges of equal magnitude and opposite signs separated by distance.

At any point at r distance from the dipole, Electric field

\(E = \frac{Kp}{r^3}\sqrt{3cos^2θ+1}\)

where p is the dipole moment, r is the distance from the dipole, θ is angle, and K is the constant.

EXPLANATION:

At any point at r distance from the dipole, Electric field

\(E = \frac{Kp}{r^3}\sqrt{3cos^2θ+1}\)

E α 1/r³

So the correct answer is option 3.

CONCEPT:

Electric field lines:

• An electric field line is an imaginary line along which a positive test charge will move if left free.
• Electric field lines are drawn to represent the electric field.

Properties of electric field lines:

• Electric field lines start from positive charges and end at negative charges. If there is a single positive charge then electric field lines start from positive charge and end at infinity. Similarly, if there is a single negative charge then electric field lines start from infinity and end at a negative charge.
• In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.

• The tangent at any point on the electric field line gives the direction of the electric field at that point.
• Electric field lines due to a point charge never intersect each other.
• The electric field line never forms a closed loop.
• The density of the electric field lines at a point indicates the strength of the electric field at that point.

EXPLANATION:

• The tangent at any point on the electric field line gives the direction of the electric field at that point.
• Electric field lines due to a point charge never intersect each other.
• If two electric field lines due to a point charge intersect with each other, then two tangents can be drawn at that point in two different directions which shows two different direction of the electric field at that point that is not possible, because at one point in the space the electric field will have only one direction. Hence, option 1 is correct.

CONCEPT:

• Gauss’s Law: Total electric flux though a closed surface is 1/εo times the charge enclosed in the surface i.e. \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\)
• But we know that Electrical flux through a closed surface is \(\oint \vec E \cdot \overrightarrow {ds} \)

\(\therefore\oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\)

Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.

EXPLANATION:

• Gauss’s Law is used to determine electric field when electric charge is continuously distributed on an object which possesses symmetrical geometry.
• The object can be a plane, cylinder, sphere, etc.
• The theorem relates electric flux associated with an electric field enclosing a symmetrical surface to the total charge enclosed by the symmetrical surface. Therefore option 1 is correct.

CONCEPT:

Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e. \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\)

But we know that Electrical flux through a closed surface is \(\oint \vec E \cdot \overrightarrow {ds} \)

\(\therefore\oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\)

Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.

EXPLANATION:

Electric field due to line charge:

Electric field due to an infinitely long straight conductor is 

\(E=\frac{\lambda }{2\pi {{\epsilon }_{o}}r}~\)

Where λ = linear charge density, r = radius of the cylinder, and εo = permittivity of free space.

From the above equation, it is clear that the electric field of an infinitely long straight wire is proportional to 1/r. Hence option 1 is correct.

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