Electric Field MCQ Quiz in मल्याळम - Objective Question with Answer for Electric Field - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

• Electric field is uniform.

• Its magnitude is 50 V/m

• It makes an angle of 30° with a positive x-axis.

OA = 6 m and OB = 8 m

Concept:

To find potential differences from the electric field, we use the relation between electric field and potential difference.

Formula:

Calculations:

∵ OA = 6 m and OB = 8 m

⇒ A = (-6 , 0) and B = (0 , 8)

Writing electric field in vector form :

where E = 50 V/m          (Magnitude of the electric field)

and 

Putting values,

∴ 

On solving and putting proper limits -

V_B - V_A = -50(3√3 + 4)

Similarly for V0 - VA :

On solving:

V0 - VA = -150√3

Concept:

By Gauss's Law, the electric field is zero inside a conducting sphere.

The electric field outside the sphere is given by:

Note: This is the same as a point charge.

Q is the charge on the sphere

d is the distance from centre outside the sphere

K is constant.

Calculation:

Given: Q = 3nC = 3 × 10^-9 C; d = 3 cm = 0.03 m; K = 9 × 10⁹ (constant)

The electric field at a distance of 3 cm will be:

Gauss' Law: This law gives the relation between the distribution of electric charge and the resulting electric field.

According to this law, the total charge Q enclosed in a closed surface is proportional to the total flux ϕ  enclosed by the surface.

ϕ ∝ Q

The Gauss law formula is expressed by:

ϕ = Q / ϵ₀

Explanation:

The electric field outside a conducting sphere

here we have Q as the charge and r as the distance.

Given: Q = 3.2 × 10−7 C 

r = 15 cm

The electric field is written as;

⇒ 

⇒ E = 0.128 × 106

⇒ E = 1.28 × 105 N/C

Hence, option 4) is the correct answer.

CONCEPT:

• Force one a charge q in an electric field is defined as

⇒ F = Eq

• The work done by this force in moving the charge by a distance d along the direction of electrostatic force (along the electric field) is

⇒ W = F.d = Eq.d

• The work done by external force in moving this charge by a distance d along the direction of the electrostatic field is

⇒ Wext = - W = - Eq.d

• By definition, the change in potential energy in moving the charge in an electrostatic field is equal to the work done by external forces

• 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

• Therefore the relation between electric potential and electric potential energy is given by

EXPLANATION:

• One electron volt (eV) is defined as a unit of energy equal to the work done on an electron in accelerating it through a potential difference of one volt.

⇒ 1 eV = 1.6 × 10-19 C × 1 V = 1.6 × 10-19 J

• Therefore option 1 is correct.

The relation between an electric field and an electric potential is given by:

E = V/r

V = Electric Potential (in Volts)

r = distance (in m)

The SI unit of E = SI unit of V/ SI unit of r = Volt/meter = V/m

Hence the SI unit of the electric field (E) = V/m

Additional Information

Electric field (E): The region around an electric charge in which other charge particles experience an attractive or repulsive force is called an electric field.

The electric field due to a charged particle is given by:

E = KQ/r2 .........(i)

Where K is constant, Q is charge and r is the distance from the charge

Electric Potential: The amount of work done in moving a unit charge from a reference point to a specific point in the electric field without producing any acceleration is called electric potential.

The electric potential due to a charged particle is given by:

V = KQ/r    ..........(ii)

Where K is constant, Q is charge and r is the distance from the charge

Volt (V) is the SI unit of the Electric Potential.

Concept:

• Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles.
• Where,  = 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 
• Magnitude of  when angle between E₁ and E₂ is 90° 
• Distance between two points A (x₁, y₁) and B (x₂, y₂) is equal to 

Calculation:

Given charge A = -0144 × 10^-9 C, B = 0.256 × 10^-9 C

Let the point (-16 cm,12 cm) is P,

Distance between points A (-16 cm, 0 cm) and P (-16 cm,12 cm), r_A =  cm

Electric field due to charge A on point P is

 = -90

Now, distance between points (0 cm, 12 cm) and P (-16 cm,12 cm), r_B =  cm

Electric field due to charge B on point P is

 = 90

Also angle between E_A and E_B is 90° then,

The magnitude of electric field 

 =127.279 ≈ 127 N/C

The correct answer is option 1

CONCEPT:

• Electric field due to a uniformly charged sphere:

Consider a non-conducting sphere of radius R with a charge Q placed inside it.

To find the electric field (E) due to the sphere at a distance r from the centre:​

Case 1: r > R (outside the sphere ) 

⇒           (sphere acts similar to point charge)

Case 2: r

⇒ 

EXPLANATION:

For a distance r that is between the centre of the sphere and radius (r

• Therefore, the linear dependency of the electric field with the distance r will be represented by a straight line through the origin. 

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:

q = 3.5 μC = 3.5 × 10^-6 C

F = 8.5 × 10^-4 N

 = 2.43 × 10² NC^-1 Answer

• Hence option 1 is the correct answer.

CONCEPT:

• Electrical charges are equally distributed on the surface of the hollow sphere.

• Inside the hollow sphere, charge distribution is done in two cases-
  • Measure at the center: All the fields are acting inside the sphere perpendicular to the surface. So, they cancel out each other and E = 0.

  •  Measure at any point other than center: In this scenario, the field acting inside is not equal, so they do not cancel each other. So, E ≠ 0.

EXPLANATION:

•  Inside the hollow sphere, the electric field is zero. Because E is a vector quantity so that they cancel out each other.

• At the center no electric field is present. Because the internal field cancels the external field. 

• Electrical charge is always equally distributed over the evenly shaped conductor.

Concept:

• An electric field is a region around a charge where other charges can feel force
•  --- (1)
• k = Coulomb' constant = 9 × 10⁹, q = charge, r = distance from charge particle
•  --- (2)
• Where ds = surface area, ϵ₀ = free space permittivity.
• q_enclosed = λl, λ = linear charge density, l = length
• Curved surface area of cylinder = 2πr

Explanation:

From equation 2

So, the magnitude of electric field in its annular region Varies as , where r is the distance from its axis.

Alternate Method

Since the charge on cylindrical capacitor present on the surface, so q = λ × r

From equation 1

So, the magnitude of electric field in its annular region Varies as , where r is the distance from its axis.