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

The correct answer is option 2) i.e. No electric field lines of force may begin or end on the same conductor

CONCEPT:

• Electric field lines: An electric field line is defined as an imaginary curve drawn in such a way that the tangent to it at each point gives the direction of the net electric field at that point. They are the visual representation of the electric field as lines.
• Equipotential surface: The surface on which all points are at the same electric potential is known as the equipotential surface.
• Electric potential: The electric potential is the difference in potential energy per unit charge between two points in an electric field. 

The electric potential (V) at a distance r from a point charge Q is given as:

​\(V = \frac{kQ}{r}\)

Where k is Coulomb's constant.

EXPLANATION:

• The line joining points of the same potential is called an equipotential line. Equipotential lines are perpendicular to electric field lines.
• The tangent to any point on the electric field line represents the direction of the electric field at that point.

Explanation:

The electric field outside a conducting sphere

\(E = \frac{1}{{4\pi { \in _0}}}\frac{Q}{{{r^2}}} \)

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;

\(E = \frac{1}{{4\pi { \in _0}}}\frac{Q}{{{r^2}}}\)

⇒ \( E= \frac{{9 × {{10}^9} × 3.2 × {{10}^{ - 7}}}}{{225 × {{10}^{ - 4}}}}\)

⇒ E = 0.128 × 106

⇒ E = 1.28 × 105 N/C

Hence, option 4) is the correct answer.

Concept:

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

The electric field outside the sphere is given by:

^{\(E = \frac{kQ}{d^2} \)} 

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:

\(E=\frac {9 × 10^9 × 3 × 10^{-9}}{0.03^{2}}=3\times 10^4V/m\)

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

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

\(⇒ \bigtriangleup U = W_{ext} = -Eq.d\)

• 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

\(⇒ V=\frac{W_{ext}}{q}\)

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

\(⇒ \bigtriangleup U = W_{ext} = Vq\)

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.

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 = -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 = \(\sqrt {(-16 - (-16))^2 + (0-12)^2} = 12\) cm

Electric field due to charge A on point P is

\(E_A = \frac{9 \times ​​10^9 \times(-0.144 \times 10^{-9})}{0.12^2}\) = -90

Now, distance between points (0 cm, 12 cm) and P (-16 cm,12 cm), r_B = \(\sqrt {(0 - (-16))^2 + (12-12)^2} = 16\) cm

Electric field due to charge B on point P is

\(E_B = \frac{9 \times ​​10^9 \times0.256 \times 10^{-9}}{0.16^2}\) = 90

Also angle between E_A and E_B is 90° then,

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

\(\Rightarrow E =\sqrt{(-90)^2 +90^2} \) =127.279 ≈ 127 N/C

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.

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 ) 

⇒ \(E= \frac {kQ}{ r^2}\)          (sphere acts similar to point charge)

Case 2: r < R and r = R (inside/surface of the sphere)

⇒ \(E = \frac {kqr}{ R^3}\)

EXPLANATION:

For a distance r that is between the centre of the sphere and radius (r < R), E ∝ 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

\(E = \frac{F}{q} = \frac{8.5 × 10^{-4}N}{3.5 × 10^{-6}C}\) = 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
• \(E = \frac{kq}{r^2}\) --- (1)
• k = Coulomb' constant = 9 × 10⁹, q = charge, r = distance from charge particle
• \(\int E\cdot ds = \frac{q_{enclosed}}{ϵ_0}\) --- (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

\(\int E\cdot ds = \frac{q_{enclosed}}{ϵ_0}\)

\(\Rightarrow E \cdot 2π rl= \frac{\lambda l}{ϵ_0}\)

\(\Rightarrow E = \frac{\lambda }{ϵ_02π r}\)

\(⇒ E \propto \frac1r\)

So, the magnitude of electric field in its annular region Varies as \(\frac{1}{r}\), 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

\(E = \frac{k\times λ \times r}{r^2} =\frac{k\times λ }{r}\)

\(⇒ E \propto \frac1r\)

So, the magnitude of electric field in its annular region Varies as \(\frac{1}{r}\), where r is the distance from its axis.