Electric Field MCQ Quiz - Objective Question with Answer for Electric Field - Download Free PDF

Explanation:

Inside a conductor, the electric field is zero due to electrostatic shielding.

Electric field lines must begin from the positive charge and terminate perpendicularly on the inner surface of the conductor.

Outside the shell, the field lines emerge perpendicularly from the outer surface, behaving as if the entire charge resides on the outer surface.

The correct field pattern must show:

• No field lines within the shell material
• Radial outward lines from the outer surface
• Lines from the central charge ending perpendicularly on the inner shell surface

Calculation:
The electric field along the axis of a uniformly charged ring is given by:

E(z) = (KQz) / (z² + R²)^3/2

To find the position where electric field is maximum, differentiate with respect to z and set dE/dz = 0:

dE/dz = 0 ⇒ z = R / √2

Given: Radius R = a√2

⇒ z = (a√2) / √2 = a

Hence, the correct option is the third one: a

Explanation:

The given equation for potential energy (E_P) is:

E_P = (K × 10) / 2² - (K × 40) / (x₀ - 2)² = 0

⇒(K × 10) / 4 - (K × 40) / (x₀ - 2)² = 0

⇒(K × 40) / (x₀ - 2)² = (K × 10) / 4

⇒40 / (x₀ - 2)² = 10 / 4

⇒x₀ - 2 = 4

So, x₀ = 6 cm

Final Answer: x₀ = 6 cm

Calculation:

The velocity of particle is given by

0.5e = (1/2)mv_x² → v_x = √(e/m)

Along x: L = v_x t = √(e/m) × t

Along y: v_y = (eE/m) × t

Dividing: v_y/L = E√(e/m) = E × v_x

⇒ tan θ = v_y/v_x = E × L = 10 × 0.1 = 1

⇒ θ = 45°

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

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:

Electric field intensity: 

• It is defined as the force experienced by a unit positive test charge in the electric field at any point.
• The positive charge experiences a force in the direction of the electric field and the negative charge experiences a force in the direction opposite of the electric field.

\(⇒ E=\frac{F}{q_{o}}\)    

Where E = electric field intensity, qo = charge on the particle

CALCULATION:

Given q_o = -6C, and F = 60 N towards north

• We know that the electric field intensity is given as,

\(⇒ E=\frac{F}{q_{o}}\)

\(⇒ E=\frac{60}{6}\)

⇒ E = 10 N/C

• The negative charge experiences a force in the direction opposite of the electric field.
• So the direction of the electric field is towards the south. Hence, option 2 is correct.

The correct answer is option 2) i.e. Electric field

CONCEPT: 

• Electric field: The region around an electric charge in which it can influence other charges is known as the electric field. 
  • The electric field intensity (E) at any point is defined as the force experienced by a unit charge placed at that location.

It is given by:

​\(E = \frac{F}{q}\)

EXPLANATION:

The electric field intensity at a point is given by:

 \(E = \frac{F}{q}\) .

• Therefore, the force acting per unit positive test charge at a location is a measure of the intensity of the electric field.

Additional Information

• Electric potential is the difference in potential energy per unit charge between two points in an electric field.
• Coulomb force is the force of interaction between electric charges separated by a distance.
• Gravity is the universal force of attraction acting between all matter. 

CONCEPT:

The electric field in the form of potential can be written as;

\(E = - \frac{{dV}}{{dr}} \) ---- (1)

Where V is the potential.

CALCULATION:

Given :

V =  0.2 m³, V = 5 V

using equation (1) we have,

\(E = - \frac{{dV}}{{dr}} \)

⇒ \(E = - \frac{{d(5)}}{{dr}} \)

⇒ E = 0

Hence, option 3) is the correct answer.

CONCEPT:

• Electric field force: When a charge moves in an electric field, a force will act on it, that is known as electric field force.
  • Due to the electric field, Force is always in the direction of the electric field if the charge is positive.

F = q E

where F is the force due to the electric field, q is the charge, and E is the electric field. 

CALCULATION:

Given that:

E = 104 Newton per Coulomb and q = 1.6 × 10^-19 C

F = q E

F = (1.6 × 10-19) × 104

F = 1.6 × 10-15 Newton

So the correct answer is option 4.

CONCEPT:

Electric field intensity: 

• It is defined as the force experienced by a unit positive test charge in the electric field at any point.

\(⇒ E=\frac{F}{q_{o}}\)    

Where E = electric field intensity, q_o = charge on the particle

EXPLANATION:

• The magnitude of electric force experienced by a charged particle in an electric field is given as,

\(\Rightarrow F=Eq_{o}\)   

• From the above equation, it is clear that the magnitude of electric force experienced by a charged particle in an electric field depends on the magnitude of the charge on the particle.
• Hence, option 1 is correct.

Concept:

Electric Field

• The electric field or 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

\(⇒ E = \frac{{kq}}{{{r^2}}}\)

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

Electric Potential

• Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field.

\(⇒ V=\frac{W}{q}\)

• The difference in electric potential between two points is called potential difference or voltage.
• The voltage is the electric field divided by the distance.

Explanation:

• The electric field is the force by charge or potential multiplied by the distance.
• SI Unit of electric potential is volt and SI unit of distance is meter.
• Unit of potential by distance is Volt / meter or V m^-1.

So, the correct option is Vm-1. 

CONCEPT:

• The space or region around the electric charge in which electrostatic force can be experienced by other charged particle is called as an electric field by that electric charge.
• The imaginary lines which is used to represent the electric field are called an electric field line.
• The tangent line at a point on the electric field line gives the direction of the electric field at that point.

• The field lines emerge from positive charge and terminate at a negative charge.
• They originate and end at right angles to the surface of the charge.
• Electric field lines do not make loop. 
• The magnitude of the electric field will be maximum where the number of field lines is maximum. 

Explanation:

From the above explanation, we can see that electric lines of force are imaginary lines used to describe the strength of an electric field.

Whereas unlike magnetic field lines, electric field lines don't form a closed loop 

Hence option 4 is incorrect among  all

Concept:

• Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles.
• ​​\(E = \frac{KQ}{r^2}\)
• Where, \(K = \frac{1}{4\pi \epsilon _0}\) = 9 × 10⁹ Nm²C^-2, Q is the charge, r is the distance from the charge.
• According to the incoming charge’s nature, the electric field will either attract or repel the charge.
• The electric field at a point is a vector quantity 
• \(\overrightarrow E= \overrightarrow E_1 + \overrightarrow E_2\)
• Magnitude of \(E =\sqrt{E_1^2 + E_2 ^2} \) when angle between E₁ and E₂ is 90° 
• Distance between two points A (x₁, y₁) and B (x₂, y₂) is equal to \(\sqrt {(x_1 - x_2)^2 + (y_1 - y_2)^2} \)

Calculation:

Given charge A = 48 × 10^-12 C, B = 36 × 10^-12 C

Let the point (-16 cm,12 cm) is P,

Distance between points A (3 cm, 0 cm) and P (3 cm,4 cm), r_A = \(\sqrt {(3 - 3))^2 + (0- 4)^2} = 4\) cm

The electric field due to charge A on point P is

\(E_A = \frac{9 × ​​10^9 ×(48 × 10^{-12})}{0.04^2}= 270 ~N/C\)

Now, the distance between points (0 cm, 4 cm) and P (3 cm, 4 cm), r_B = \(\sqrt {(0 - 3))^2 + (4- 4)^2} = 3\) cm

The electric field due to charge B on point P is

\(E_B = \frac{9 × ​​10^9 ×(36 × 10^{-12})}{0.03^2}= 360~ N/C\)

Also the angle between E_A and E_B is 90° then,

The magnitude of the electric field \(E =\sqrt{E_A^2 + E_B ^2} \)

\(\Rightarrow E =\sqrt{(270)^2 +(360)^2} \) = 4.5 × 10² N/C

Explanation:

Given

We know that,

Electric field due to the uniformly charged large surface is given as:

E = σ/ 2ϵ_o  ----- (1)

From equation (1) it is clear that the electric field due to a uniformly charged surface is independent of distance from the surface.

It only depends on the surface density.

Hence in both cases, the Electric field will be the same, i.e.

E₁ = E2 = σ/ 2ϵo

Hence option 3) is the correct choice.