Class 11 - Unit # 08 : Electric Fields - Solved Numericals

Physics Numerical Sheet
Electric Fields

1. (a) Calculate the value of two equal charges if they repel one another with a force of 0.1 N when situated 50 cm apart in a vacuum. (b) What would be the size of the charges if they were situated in an insulating liquid whose permittivity was ten times that of a vacuum?

Data:

• Repulsive Force (F) = 0.1 N
• Distance (r) = 50 cm = 0.5 m
• Coulomb's constant in vacuum (k) = 9 × 10⁹ N·m²/C²
• Relative permittivity of liquid (ε_r) = 10

Solution:

(a) In a vacuum: According to Coulomb's Law, for two equal charges ($q_1 = q_2 = q$):

F = (k × q²) / r²

0.1 = (9 × 10⁹ × q²) / (0.5)²

0.1 = (9 × 10⁹ × q²) / 0.25

q² = (0.1 × 0.25) / (9 × 10⁹) = 2.78 × 10⁻¹²

q = √(2.78 × 10⁻¹²) ≈ 1.67 × 10⁻⁶ C = 1.67 μC

(b) In the insulating liquid: The force constant scales down inversely by the relative permittivity:

F = (k × q'²) / (ε_r × r²)

0.1 = (9 × 10⁹ × q'²) / (10 × 0.25)

q'² = (0.1 × 2.5) / (9 × 10⁹) = 2.78 × 10⁻¹¹

q' = √(2.78 × 10⁻¹¹) ≈ 5.27 × 10⁻⁶ C = 5.27 μC

RESULT: The magnitude of each charge in a vacuum is 1.67 × 10⁻⁶ C and inside the insulating medium is 5.27 × 10⁻⁶ C.

2. How far apart must two protons be if the magnitude of the electrostatic force acting on either one due to the other is equal to the magnitude of the gravitational force on a proton at Earth's surface?

Data:

• Mass of a proton (m) = 1.67 × 10⁻²⁷ kg
• Charge of a proton (q) = 1.6 × 10⁻¹⁹ C
• Coulomb's constant (k) = 9 × 10⁹ N·m²/C²
• Acceleration due to gravity (g) = 9.8 m/s²

Solution:

According to the given condition, the Electrostatic Force ($F_e$) balances the Gravitational Force ($F_g$):

F_e = F_g

(k × q²) / r² = m × g

r² = (k × q²) / (m × g)

r² = (9 × 10⁹ × (1.6 × 10⁻¹⁹)²) / (1.67 × 10⁻²⁷ × 9.8)

r² = (9 × 10⁹ × 2.56 × 10⁻³⁸) / (1.6366 × 10⁻²⁶)

r² = (2.304 × 10⁻²⁸) / (1.6366 × 10⁻²⁶) ≈ 1.408 × 10⁻²

r = √(0.01408) ≈ 0.118 m

RESULT: The two protons must be separated by a distance of 0.118 m.

3. An electron of charge 1.6 × 10⁻¹⁹ C is situated in a uniform electric field of intensity 1200 V/cm. Find the force on it, its acceleration, and the time it takes to travel 2 cm from rest (electronic mass, m_e = 9.1 × 10⁻³¹ kg).

Data:

• Charge of electron (e) = 1.6 × 10⁻¹⁹ C
• Electric field intensity (E) = 1200 V/cm = 1200 / 10⁻² = 1.2 × 10⁵ V/m
• Mass of electron (m_e) = 9.1 × 10⁻³¹ kg
• Distance traveled (d) = 2 cm = 0.02 m
• Initial velocity (v_i) = 0 m/s (from rest)

Solution:

(a) Electric Force:

F = e × E

F = (1.6 × 10⁻¹⁹) × (1.2 × 10⁵) = 1.92 × 10⁻¹⁴ N

(b) Acceleration: According to Newton's Second Law ($F = ma$):

a = F / m_e

a = (1.92 × 10⁻¹⁴) / (9.1 × 10⁻³¹) ≈ 2.11 × 10¹⁶ m/s²

(c) Time taken: Using the 2nd Equation of Motion ($d = v_i·t + ½at²$):

d = 0 + ½ × a × t²

t = √(2d / a)

t = √(2 × 0.02 / (2.11 × 10¹⁶)) = √(1.895 × 10⁻¹⁸) ≈ 1.38 × 10⁻⁹ s

RESULT: The force on the electron is 1.92 × 10⁻¹⁴ N, its acceleration is 2.11 × 10¹⁶ m/s², and the transit time is 1.38 × 10⁻⁹ s.

4. An alpha particle (the nucleus of a helium atom) has a mass of 6.64 × 10⁻²⁷ kg and a charge of 2e. What are the (a) magnitude and (b) direction of the electric field that will balance the gravitational force on the particle?

Data:

• Mass of alpha particle (m) = 6.64 × 10⁻²⁷ kg
• Charge of alpha particle (q) = 2e = 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
• Acceleration due to gravity (g) = 9.8 m/s²
• Electric field (E) = ?

Solution:

To balance the gravitational pull, the upward electric force must equal the downward gravitational weight:

F_e = F_g ➔ q × E = m × g

E = (m × g) / q

E = (6.64 × 10⁻²⁷ × 9.8) / (3.2 × 10⁻¹⁹)

E = (6.5072 × 10⁻²) / (3.2 × 10⁻¹⁹) ≈ 2.03 × 10⁻⁷ N/C

Since the alpha particle is positively charged, the electric force is in the same direction as the field lines. Therefore, the field must point vertically upward to oppose gravity.

RESULT: The required electric field magnitude is 2.03 × 10⁻⁷ N/C and its direction is vertically upward.

5. A proton and an electron form two corners of an equilateral triangle of side length 2.0 × 10⁻⁶ m. What is the magnitude of the net electric field these two particles produce at the third corner?

Data:

• Charge magnitude ($|q_1| = |q_2| = q$) = 1.6 × 10⁻¹⁹ C
• Side length of the triangle (r) = 2.0 × 10⁻⁶ m
• Coulomb's constant (k) = 9 × 10⁹ N·m²/C²

Solution:

The magnitude of the electric field produced by the proton ($E_+$) and electron ($E_-$) at the third vertex is identical due to symmetry:

E_+ = E_- = (k × q) / r²

E_+ = (9 × 10⁹ × 1.6 × 10⁻¹⁹) / (2.0 × 10⁻⁶)²

E_+ = (1.44 × 10⁻⁹) / (4.0 × 10⁻¹²) = 360 N/C

In an equilateral triangle, the angle between the two field vector fields at the target corner resolves such that their resultant vector fields form a parallel component across a 120-degree exterior orientation ($2 \times E \times \cos(60^\circ)$):

E_{net} = √[E_+² + E_-² + 2E_+E_-\cos(120°)]

Since $\cos(120°) = -0.5$ and $E_+ = E_- = E$:

E_{net} = √[E² + E² - E²] = E = 360 N/C

RESULT: The net electric field intensity at the third corner is 3.6 × 10² N/C.

6. Figure shows two charged particles on an x-axis: -q = -3.20 × 10⁻¹⁹ C at x = -3.00 m and q = 3.20 × 10⁻¹⁹ C at x = 3.00 m. What are the (a) magnitude and (b) direction (relative to the positive direction of the x-axis) of the net electric field produced at point P at y = 4.00 m?

Data:

• Charge magnitude (q) = 3.20 × 10⁻¹⁹ C
• Coordinates: Charge 1 (-q) at (-3, 0), Charge 2 (+q) at (3, 0), Point P at (0, 4)
• Coulomb's constant (k) = 9 × 10⁹ N·m²/C²

Solution:

1. Determine the spatial line of sight distance ($r$) from either charge to Point P using Pythagoras theorem:

r = √(x² + y²) = √(3² + 4²) = √25 = 5.00 m

2. Find the individual electric field magnitudes:

E_+ = E_- = (k × q) / r² = (9 × 10⁹ × 3.20 × 10⁻¹⁹) / 5²

E_+ = (2.88 × 10⁻⁹) / 25 = 1.152 × 10⁻¹⁰ N/C

3. Resolve into components. The vertical components ($\sin\theta$) cancel out, while the horizontal components ($\cos\theta$) reinforce each other along the negative x-axis direction:

E_{net} = 2 × E_+ × \cos(\theta)

From geometry, $\cos(\theta) = \text{base} / \text{hypotenuse} = 3.00 / 5.00 = 0.6$:

E_{net} = 2 × (1.152 × 10⁻¹⁰) × 0.6 = 1.38 × 10⁻¹⁰ N/C

RESULT: The net electric field magnitude is 1.38 × 10⁻¹⁰ N/C, directed towards the negative x-axis (180°).

7. A proton and an electron form two corners of an equilateral triangle of side length 5 μm. What is the magnitude of the net electric field these two particles produce at the third corner?

Data:

• Charge magnitude (q) = 1.6 × 10⁻¹⁹ C
• Distance separation (r) = 5 μm = 5 × 10⁻⁶ m
• Coulomb's constant (k) = 9 × 10⁹ N·m²/C²

Solution:

Calculate the base value of the symmetric fields matching the geometry from problem 5:

E = (k × q) / r²

E = (9 × 10⁹ × 1.6 × 10⁻¹⁹) / (5 × 10⁻⁶)²

E = (1.44 × 10⁻⁹) / (25 × 10⁻¹²) = 57.6 N/C

For an equilateral triangle configuration pairing a matching positive and negative node, the horizontal components combine directly to yield a net field equal to the individual field magnitude:

E_{net} = E = 57.6 N/C

RESULT: The net electric field at the third corner is 57.6 N/C.

8. The square surface shown in figure measures 3.2 mm on each side. It is immersed in a uniform electric field with magnitude E = 1800 N/C and with field lines at an angle of θ = 35° with a normal to the surface. Calculate the electric flux through the surface.

Data:

• Side of square surface (s) = 3.2 mm = 3.2 × 10⁻³ m
• Area of surface (A) = s² = (3.2 × 10⁻³)² = 1.024 × 10⁻⁵ m²
• Electric field magnitude (E) = 1800 N/C
• Angle with respect to the normal (θ) = 35°

Solution:

According to the mathematical definition of electric flux:

Φ = E × A × \cos(\theta)

Φ = 1800 × (1.024 × 10⁻⁵) × \cos(35°)

Φ = 0.018432 × 0.81915 ≈ 0.0151 N·m²/C

RESULT: The electric flux traversing through the square surface area is 1.51 × 10⁻² N·m²/C.

9. An electron is liberated from the lower of two large parallel metal plates separated by a distance h = 2 cm. The upper plate has a potential of 2400 volts relative to the lower. How long does the electron take to reach it?

Data:

• Separation distance (h) = 2 cm = 0.02 m
• Potential difference (V) = 2400 V
• Mass of electron (m_e) = 9.1 × 10⁻³¹ kg
• Charge of electron (e) = 1.6 × 10⁻¹⁹ C

Solution:

1. Find the internal uniform electric field intensity:

E = V / h = 2400 / 0.02 = 1.2 × 10⁵ V/m

2. Determine the acceleration driving the target electron charge particle:

a = (e × E) / m_e

a = (1.6 × 10⁻¹⁹ × 1.2 × 10⁵) / (9.1 × 10⁻³¹) ≈ 2.11 × 10¹⁶ m/s²

3. Solve for time using the standard distance kinematic relation starting from rest ($h = ½at²$):

t = √(2h / a)

t = √(2 × 0.02 / (2.11 × 10¹⁶)) ≈ 1.38 × 10⁻⁹ s

RESULT: The net time window required for the electron to traverse between the plates is 1.38 × 10⁻⁹ s.

10. Two large parallel metal plates are 1.5 cm apart and have charges of equal magnitudes but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then 5.0 V, what is the electric field in the region between the plates?

Data:

• Total plate spacing separation ($d_{total}$) = 1.5 cm = 0.015 m
• Midpoint baseline position ($d_{mid}$) = 1.5 / 2 = 0.75 cm = 0.0075 m
• Potential value at midpoint position ($V_{mid}$) = 5.0 V

Solution:

The electric field between parallel plates is uniform. We can use the potential gradient relationship evaluated at the halfway mark context:

E = V_{mid} / d_{mid}

E = 5.0 / 0.0075 ≈ 666.67 V/m

RESULT: The static uniform electric field in the region between the plates is 667 V/m.