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:

\(V_B-V_A = -\int_{A}^B \vec{E}.\vec{dr}\)

Calculations:

∵ OA = 6 m and OB = 8 m

⇒ A = (-6 , 0) and B = (0 , 8)

Writing electric field in vector form :

\(\vec{E} = Ecos30^\circ(\hat{i}) + Esin30^\circ(\hat{j})\)

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

and \(\vec{dr} = dx(\hat{i}) + dy(\hat{j})\)

Putting values,

∴ \(V_B-V_A = -\int_{(-6,0)}^{(0,8)} 50(\frac{√ 3}{2}dx + \frac{1}{2}dy)\)

On solving and putting proper limits -

V_B - V_A = -50(3√3 + 4)

Similarly for V0 - VA :

\(V_O-V_A = -\int_{(-6,0)}^{(0,0)} 50(\frac{√ 3}{2}dx + \frac{1}{2}dy)\)

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:

^{\(E = \frac{kQ}{d^2} \)} 

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

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:

• 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.

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.
• \(E = \frac{KQ}{r^2}\)
• 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 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.