CBSE Class 12 Physics Electric Charges And Fields Worksheet Set 02

CBSE Class 12 Physics Electric Charges And Fields Important Questions (2).Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects. 

1.Define Torque. Find the torque acting on an electric dipole placed in a uniform electric field. Give the conditions for torque acting on an electric dipole is

(a) zero and (b) maximum 

2 How can you explain a comb run through dry hair attracts pieces of paper?

3 Define linear charge density. Give its SI unit of measurement. Also give the formula to find the linear charge density. 

4 Define the following. Also give their SI units of measurement?

(a) Surface charge density (b) volume charge density. 

5 State & prove Gauss’s Law. 

6 Give any four important points regarding Gauss’s law. 
 

Important Questions for NCERT Class 12 Physics Electric Charges And Fields
 

Question. A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field vector E. Due to the force qEvector , its velocity increases from 0 to 6 m s–1 in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively 
(a) 2 m s^–1, 4 m s^–1
(b) 1 m s^–1, 3 m s^–1
(c) 1 m s^–1, 3.5 m s^–1
(d) 1.5 m s^–1, 3 m s^–1

Question. A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is   
(a) qEy
(b) qE²y
(c) qEy²
(d) q²Ey

Question. Which of the following pairs does not have similar dimensions ?
(a) tension and surface tension
(b) stress and pressure
(c) Planck’s constant and angular momentum
(d) angle and strain

Question. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will   
(a) increase four times
(b) be reduced to half
(c) remain the same
(d) be doubled

Question. A charge Q is situated at the corner of a cube, the electric flux passed through all the six faces of the cube is   
(a) Q/6ε₀
(b) Q/8ε₀
(c) Q/ε₀
(d) Q/2ε₀

Question. A point charge + q is placed at the centre of a cube of side l. The electric flux emerging from the cube is   
(a) 6ql2/ε0
(b) q/6l2ε0
(c) zero
(d) q/ε0

Question. Three point charges +q, –2q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are   
(a) 2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
(b) qa along the line joining points (x = 0, y = 0,z = 0) and (x = a, y = a, z = 0)
(c) 2qa along + x direction
(d) 2qa along + y direction. 

Question. A point Q lies on the perpendicular bisector of an electrical dipole of dipole moment p. If the distance of Q from the dipole is r (much larger than the size of the dipole), then the electric field at Q is proportional to   
(a) p² and r^–3 
(b) p and r^–2
(c) p^–1 and r^–2
(d) p and r^–3

Question. An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 10⁵ N C^–1.   
It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is
(a) 8 mC
(b) 2 mC
(c) 5 mC
(d) 7 mC

Question. Which of the following physical quantities do not have same dimensions? 
(a) pressure and stress
(b) tension and surface tension
(c) strain and angle
(d) energy and work.

Question. What is the flux through a cube of side a if a point charge of q is at one of its corner? 
(a) 2q/ε0
(b) q/8ε0
(c) q/ε0
(d) q/2ε0 . 6a²

Question. A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre   
(a) decreases as r increases for r < R and for r > R
(b) increases as r increases for r < R and for r > R
(c) zero as r increases for r < R, decreases as rincreases for r > R
(d) zero as r increases for r < R, increases as r increases for r > R

Question. The electric field at a distance 3R/2 R from the centre of a charged conducting spherical shell of radius R is E.   The electric field at a distance R/2from the centre of the sphere is
(a) zero
(b) E
(c) E/2
(d) E/3

Question. A hollow insulated conduction sphere is given a positive charge of 10 mC. What will be the electric field at the centre of the sphere if its radius is 2 metres ?   
(a) 20 mC m^–2
(b) 5 mC m^–2
(c) zero
(d) 8 mC m^–2

Question 1. Define Torque. Find the torque acting on an electric dipole placed in a uniform electric field. Give the conditions for torque acting on an electric dipole is (a) zero and (b) maximum
Answer:
1. **Definition of Torque:**
Torque is defined as the measure of a rotational force that causes an object to turn about a pivot or axis of rotation.

2. **Torque on an Electric Dipole in a Uniform Electric Field:**
Let us consider an electric dipole consisting of two charges \( +q \) and \( -q \), separated by a distance \( 2a \), placed in a uniform electric field \( \vec{E} \) such that the dipole moment \( \vec{p} \) makes an angle \( \theta \) with the direction of the field.
The electrostatic force acting on the charge \( +q \) is \( \vec{F}_1 = q\vec{E} \), which acts in the direction of the field.
The electrostatic force acting on the charge \( -q \) is \( \vec{F}_2 = -q\vec{E} \), which acts opposite to the direction of the field.
Since the two forces are equal in magnitude and opposite in direction, the net translational force on the dipole is zero:
\[ \vec{F}_{\text{net}} = q\vec{E} - q\vec{E} = 0 \]
However, because these two forces act along different lines of action, they form a couple that exerts a torque on the dipole. The magnitude of this torque \( \tau \) is equal to the product of the magnitude of either force and the perpendicular distance between their parallel lines of action:
\[ \tau = \text{Force} \times \text{Perpendicular distance} \]

\( \implies \tau = (qE) \cdot (2a \sin(\theta)) \)
Since the electric dipole moment is defined as \( p = q \cdot 2a \), we can substitute this to obtain:
\[ \tau = pE \sin(\theta) \]
In vector notation, this rotational relation is expressed as:
\[ \vec{\tau} = \vec{p} \times \vec{E} \]

3. **Conditions for Torque:**
(a) **Zero Torque (\( \tau = 0 \)):**
Torque is zero when \( \sin(\theta) = 0 \), which occurs at:
- \( \theta = 0^\circ \) (The dipole is aligned parallel to the field, representing a state of stable equilibrium).
- \( \theta = 180^\circ \) (The dipole is aligned antiparallel to the field, representing a state of unstable equilibrium).

(b) **Maximum Torque (\( \tau = \tau_{\text{max}} \)):**
Torque is maximum when \( \sin(\theta) = 1 \), which occurs at:
- \( \theta = 90^\circ \) (The dipole is aligned perpendicular to the direction of the electric field).
The maximum torque value is:
\[ \tau_{\text{max}} = pE \]
In simple words: Torque is the turning force that rotates the dipole. When placed in an electric field, the opposite forces on its two ends make it twist until it aligns with the field. This twisting force is zero when it is already aligned and reaches its maximum when it is perpendicular to the field.

Exam Tip: Always specify the direction of the forces on both charges and explicitly define the perpendicular distance (\( 2a \sin(\theta) \)) using a quick geometry step to score full marks.

Question 2. How can you explain a comb run through dry hair attracts pieces of paper?
Answer: When a plastic comb is run through dry hair, it acquires a net static electric charge due to friction. When this charged comb is brought near neutral pieces of paper, it polarizes the molecules within the paper. This polarization induces opposite charges on the closer edge of the paper and similar charges on the farther edge. Because the opposite, attractive charges are closer to the comb than the similar, repulsive charges, the net electrostatic force is attractive, causing the paper pieces to be drawn toward the comb.
In simple words: Rubbing the comb through dry hair charges it up. When brought near paper, this charge pulls the opposite charges in the paper closer, creating an attractive force that lifts the paper.

Exam Tip: Be sure to use technical terms such as "frictional charging" and "polarization of neutral molecules" to explain the mechanism clearly.

Question 3. Define linear charge density. Give its SI unit of measurement. Also give the formula to find the linear charge density.
Answer: Linear charge density is defined as the amount of electric charge distributed per unit length along a one-dimensional linear conductor.
The formula to calculate the linear charge density \( \lambda \) is:
\[ \lambda = \frac{q}{L} \]
where:
- \( q \) is the total charge.
- \( L \) is the total length of the conductor.
The SI unit of measurement for linear charge density is **Coulomb per meter** (\( \text{C m}^{-1} \) or \( \text{C/m} \)).
In simple words: Linear charge density tells us how much electric charge is spread out along each meter of a thin wire. It is found by dividing the total charge by the wire's length.

Exam Tip: Always state both the formula and the correct SI unit clearly, and specify what each variable represents in your formula.

Question 4. Define the following. Also give their SI units of measurement?
(a) Surface charge density (b) volume charge density.
Answer:
(a) **Surface Charge Density (\( \sigma \)):**
It is defined as the quantity of electric charge distributed per unit surface area over a two-dimensional surface. The formula is:
\[ \sigma = \frac{q}{A} \]
where \( q \) is the charge and \( A \) is the surface area. The SI unit of measurement is **Coulomb per square meter** (\( \text{C m}^{-2} \) or \( \text{C/m}^2 \)).

(b) **Volume Charge Density (\( \rho \default \)):**
It is defined as the quantity of electric charge distributed per unit volume within a three-dimensional object. The formula is:
\[ \rho = \frac{q}{V} \]
where \( q \) is the charge and \( V \) is the volume. The SI unit of measurement is **Coulomb per cubic meter** (\( \text{C m}^{-3} \) or \( \text{C/m}^3 \)).
In simple words: Surface charge density measures charge spread over a flat area, while volume charge density measures how crowded charge is inside a solid 3D space.

Exam Tip: Write down the mathematical formulas and define the variables alongside their SI units to ensure you meet all marking requirements.

Question 5. State & prove Gauss’s Law.
Answer:
1. **Statement:**
Gauss's Law states that the net total electric flux \( \Phi \) passing through any closed imaginary surface (known as a Gaussian surface) is equal to \( \frac{1}{\varepsilon_0} \) times the net charge \( q \) enclosed within that surface. Mathematically:
\[ \Phi = \oint \vec{E} \cdot d\vec{S} = \frac{q}{\varepsilon_0} \]

2. **Proof:**
Let us place a positive point charge \( q \) at the center of a sphere of radius \( r \). By spherical symmetry, the electric field \( \vec{E} \) at any point on the surface of this sphere is directed radially outwards, and its magnitude is:
\[ E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \]
Consider a tiny area element \( d\vec{S} \) on the surface of the sphere. The area vector \( d\vec{S} \) also points radially outwards, perpendicular to the surface. Thus, the angle \( \theta \) between \( \vec{E} \) and \( d\vec{S} \) is \( 0^\circ \).
The electric flux \( d\Phi \) through this small area element is:
\[ d\Phi = \vec{E} \cdot d\vec{S} = E \cdot dS \cos(0^\circ) = E \cdot dS \]
The total electric flux \( \Phi \) through the entire spherical surface is found by integrating this expression over the closed surface:
\[ \Phi = \oint \vec{E} \cdot d\vec{S} = \oint E \cdot dS \]
Since the magnitude of the electric field \( E \) is constant at all points on the sphere's surface, we can pull it out of the integral:
\[ \Phi = E \oint dS \]
The total surface area of the sphere is \( \oint dS = 4\pi r^2 \). Substituting the expressions for \( E \) and the total area:

\( \implies \Phi = \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \right) \cdot (4\pi r^2) \default \)

\( \implies \Phi = \frac{q}{\varepsilon_0} \)
This completes the proof of Gauss's Law.
In simple words: Gauss's Law states that the total electric field lines passing out of a closed surface depend only on the charge trapped inside. We prove this by calculating the field on a sphere and showing that the radius cancels out completely.

Exam Tip: When proving Gauss's Law, make sure to state that the angle between the electric field vector and the area vector is \( 0^\circ \), which simplifies the dot product to a direct scalar multiplication.

Question 6. Give any four important points regarding Gauss’s law.
Answer: Four key points regarding Gauss's Law are:
1. The law is valid and applicable for any closed surface of arbitrary shape or size.
2. The charge term \( q \) in the formula represents only the sum of charges enclosed inside the surface; any charge located outside the surface does not contribute to the net flux.
3. The positions of the enclosed charges within the closed surface do not affect the total electric flux.
4. The imaginary closed surface chosen to calculate the flux is called a Gaussian surface, and it should not pass through any discrete point charge because the electric field at the exact location of a point charge is not uniquely defined.
In simple words: Gauss's Law works for any closed shape, only counts charges on the inside, doesn't care where those charges are located, and requires an imaginary surface that avoids intersecting individual point charges.

Exam Tip: Listing these conceptual points clearly in numbered lists is highly recommended to ensure the examiner can easily award full marks.

Question 7. The electric field components in fig are Ex= αx ½. Ey=Ez=0, in which α=800N/Cm ½. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a= 0.1m.
Answer:
The given electric field is directed along the X-axis only, as \( E_y = E_z = 0 \). Therefore, the electric flux is non-zero only through the left and right faces of the cube, which are perpendicular to the X-axis. The flux through the remaining four faces is zero because the electric field is parallel to them.

Let the side of the cube be \( a = 0.1 \text{ m} \).
- **On the Left Face (at \( x = a \)):**
The electric field is \( E_L = \alpha a^{1/2} \). The outward normal vector \( \hat{n}_L \) points along \( -\hat{i} \), while \( \vec{E} \) points along \( +\hat{i} \). Thus, the angle is \( 180^\circ \).
\[ \Phi_L = E_L \cdot A \cos(180^\circ) = -\alpha a^{1/2} \cdot a^2 = -\alpha a^{5/2} \default \]
- **On the Right Face (at \( x = 2a \)):**
The electric field is \( E_R = \alpha (2a)^{1/2} \). The outward normal vector \( \hat{n}_R \) points along \( +\hat{i} \). Thus, the angle is \( 0^\circ \).
\[ \Phi_R = E_R \cdot A \cos(0^\circ) = \alpha \sqrt{2} a^{1/2} \cdot a^2 = \alpha \sqrt{2} a^{5/2} \default \]

(a) **Total Flux through the Cube (\( \Phi \)):**
The net flux through the entire cube is the sum of the fluxes through the left and right faces:
\[ \Phi = \Phi_L + \Phi_R = \alpha a^{5/2} (\sqrt{2} - 1) \]
Given \( \alpha = 800 \text{ N/(C m}^{1/2}\text{)} \) and \( a = 0.1 \text{ m} \):

\( \implies a^{5/2} = (0.1)^{5/2} \approx 0.003162 \text{ m}^{5/2} \default \)

\( \implies \Phi = 800 \cdot 0.003162 \cdot (1.414 - 1) \)

\( \implies \Phi \approx 2.53 \cdot 0.414 \approx 1.05 \text{ N m}^2/\text{C} \)

(b) **Total Charge within the Cube (\( q \)):**
According to Gauss's Law, the charge \( q \) is related to the net flux by:
\[ q = \Phi \cdot \varepsilon_0 \]

\( \implies q = (1.05 \text{ N m}^2/\text{C}) \cdot (8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2) \)

\( \implies q \approx 9.27 \times 10^{-12} \text{ C} \)
In simple words: Since the electric field only flows sideways, flux only enters and exits the left and right sides. Calculating the fields on both sides gives a net flux of \( 1.05 \text{ N m}^2/\text{C} \), which corresponds to a trapped charge of \( 9.27 \times 10^{-12} \text{ C} \) inside the cube.

Exam Tip: Be careful with fractional exponents when calculating \( a^{5/2} \) and remember that the area vectors on opposite sides point in opposite directions, changing the signs of the flux components.

Question 8. Derive an expression to find the electric field due to an infinitely long thin straight wire using Gauss’s Law
Answer:
Consider an infinitely long, straight, thin wire carrying a uniform linear charge density \( \lambda \). By cylindrical symmetry, the electric field \( \vec{E} \) is directed radially outwards, perpendicular to the wire, and its magnitude depends only on the radial distance \( r \).
To find the electric field at a distance \( r \), we construct a coaxial cylindrical Gaussian surface of radius \( r \) and length \( l \). This cylindrical surface consists of three sections:
1. **Top Flat Cap (\( S_1 \)):** The area vector \( d\vec{S}_1 \) points vertically upwards, while the electric field \( \vec{E} \) points horizontally outwards. The angle between them is \( 90^\circ \), so the flux is zero.
2. **Bottom Flat Cap (\( S_2 \default \)):** The area vector \( d\vec{S}_2 \) points vertically downwards, and \( \vec{E} \) is horizontal. The angle is \( 90^\circ \), so the flux is zero.
3. **Curved Cylindrical Surface (\( S_3 \)):** At any point on this surface, the area vector \( d\vec{S}_3 \) points radially outwards, parallel to the electric field \( \vec{E} \). The angle is \( 0^\circ \).

The total electric flux \( \Phi \) through the cylindrical Gaussian surface is:
\[ \Phi = \oint \vec{E} \cdot d\vec{S} = \int_{S_1} E \, dS \cos(90^\circ) + \int_{S_2} E \, dS \cos(90^\circ) + \int_{S_3} E \, dS \cos(0^\circ) \]

\( \implies \Phi = 0 + 0 + E \int_{S_3} dS \default \)
Since the magnitude of the field \( E \) is constant everywhere on the curved surface of area \( 2\pi r l \):
\[ \Phi = E \cdot (2\pi r l) \]
The net charge enclosed within this cylindrical Gaussian surface of length \( l \) is:
\[ q_{\text{enclosed}} = \lambda l \]
Applying Gauss's Law:
\[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} \]

\( \implies E \cdot (2\pi r l) = \frac{\lambda l}{\varepsilon_0} \default \)
Canceling \( l \) from both sides yields the magnitude of the electric field:
\[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]
In vector notation, this is written as:
\[ \vec{E} = \frac{\lambda}{2\pi \varepsilon_0 r} \hat{r} \]
where \( \hat{r} \) is the unit vector directed radially outwards from the wire.
In simple words: To find the field of a long wire, we place a cylinder around it. The field only passes out through the curved sides of the cylinder. By equating this total passing field to the enclosed charge, we find that the field decreases inversely with the distance.

Exam Tip: Be sure to write down the individual integral terms for all three parts of the cylindrical surface to show a complete, rigorous proof.

Question 9. Derive an expression to find the electric field due to a uniformly charged infinite plane sheet using Gauss’s Law
Answer:
Consider an infinite plane thin sheet of charge with a uniform surface charge density \( \sigma \). By planar symmetry, the electric field \( \vec{E} \) must be perpendicular to the sheet, pointing outward away from it on both sides. The magnitude of the electric field is identical at equal distances on either side of the sheet.

To find the field at a distance \( r \) from the sheet, we construct a cylindrical Gaussian surface of cross-sectional area \( A \) and length \( 2r \), positioned perpendicular to the sheet so that the sheet cuts the cylinder symmetrically in half.
The surface of this cylinder consists of:
1. **Curved Surface:** The area vector at any point on the curved surface is perpendicular to the electric field lines (angle \( 90^\circ \)). Thus, the flux through the curved surface is zero.
2. **Two Flat circular Caps:** At both end caps (each of area \( A \)), the electric field lines point outwards, parallel to the area vector (angle \( 0^\circ \)).

The total electric flux \( \Phi \) through the cylindrical Gaussian surface is:
\[ \Phi = \int_{\text{Left End}} \vec{E} \cdot d\vec{S} + \int_{\text{Right End}} \vec{E} \cdot d\vec{S} + \int_{\text{Curved}} \vec{E} \cdot d\vec{S} \default \]

\( \implies \Phi = E \cdot A + E \cdot A + 0 = 2EA \)
The total charge enclosed within this Gaussian cylinder of cross-section \( A \) is:
\[ q_{\text{enclosed}} = \sigma A \]
Applying Gauss's Law:
\[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} \]

\( \implies 2EA = \frac{\sigma A}{\varepsilon_0} \default \)
Canceling the cross-sectional area \( A \) from both sides yields:
\[ E = \frac{\sigma}{2\varepsilon_0} \]
This shows that the electric field near a uniformly charged infinite plane sheet is constant and independent of the distance from the sheet.
In simple words: By sliding a cylinder through a flat charged sheet, we find that the field lines only push out of the two flat ends. Dividing this total escaping field by the trapped charge reveals that the field is uniform and does not fade with distance.

Exam Tip: State clearly that the electric field is completely independent of the distance \( r \) from the sheet, as this is a key conceptual conclusion of the derivation.

Question 10. Derive an expression to find the electric field due to a uniformly charged thin spherical shell using Gauss’s Law
Answer:
Let us analyze a thin spherical shell of radius \( R \) carrying a total charge \( q \) uniformly distributed over its surface. By spherical symmetry, the electric field \( \vec{E} \) must be radial in direction, pointing outwards (if \( q > 0 \)). To find the electric field at any point at a distance \( r \) from the center, we choose a concentric spherical Gaussian surface of radius \( r \). At any point on this surface, the angle between \( \vec{E} \) and \( d\vec{S} \) is \( 0^\circ \).
The total electric flux \( \Phi \) through this Gaussian surface is:
\[ \Phi = \oint \vec{E} \cdot d\vec{S} = E \oint dS = E \cdot (4\pi r^2) \default \]

- **Case 1: Outside the shell (\( r > R \default \)):**
The spherical Gaussian surface completely encloses the charged shell, so the enclosed charge is:
\[ q_{\text{enclosed}} = q \]
Applying Gauss's Law:
\[ E \cdot (4\pi r^2) = \frac{q}{\varepsilon_0} \]

\( \implies E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \default \)
This shows that for any point outside, the shell acts as if its entire charge were concentrated at its center.

- **Case 2: Inside the shell (\( r < R \default \)):**
The Gaussian surface of radius \( r \) lies entirely inside the shell, meaning there is no charge enclosed within it:
\[ q_{\text{enclosed}} = 0 \]
Applying Gauss's Law:
\[ E \cdot (4\pi r^2) = 0 \]

\( \implies E = 0 \default \)
Thus, the electric field is zero at all points inside a charged conducting spherical shell.
In simple words: Outside a charged sphere, the electric field acts normally, as if all the charge were squished into a single point at the center. Inside the sphere, there is no trapped charge, so the electric field drops to exactly zero.

Exam Tip: Be sure to write down both cases (inside and outside) separately to secure full marks for this derivation.

Question. Two charges ±10μC are placed 5.0 mm apart. Determine the electric field at (a) a point on the axis of the dipole at 15 cm away from the center of the dipole on the axial line and (b) at 15 com away from the center of the dipole on the equatorial line of the dipole.
Answer:
Given:
- Magnitude of charge, \( q = 10 \text{ }\mu\text{C} = 10 \times 10^{-6} \text{ C} \)
- Separation distance, \( 2a = 5.0 \text{ mm} = 5.0 \times 10^{-3} \text{ m} \default \)
- Distance from the center, \( r = 15 \text{ cm} = 0.15 \text{ m} \default \)
- Dipole moment, \( p = q \cdot 2a = (10 \times 10^{-6} \text{ C}) \cdot (5.0 \times 10^{-3} \text{ m}) = 5.0 \times 10^{-8} \text{ C m} \)

Since the distance \( r = 15 \text{ cm} \) is much larger than the separation \( 2a = 0.5 \text{ cm} \), we can use the short-dipole approximations:

(a) **At a point on the axial line:**
The electric field along the axial line of a short dipole is:
\[ E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3} \]
Substituting the values:

\( \implies E_{\text{axial}} = \frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2) \cdot 2 \cdot (5.0 \times 10^{-8} \text{ C m})}{(0.15 \text{ m})^3} \default \)

\( \implies E_{\text{axial}} = \frac{900}{0.003375} \approx 2.67 \times 10^5 \text{ N/C} \)
The direction of this field is along the direction of the dipole moment vector.

(b) **At a point on the equatorial line:**
The electric field along the equatorial line of a short dipole is:
\[ E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} = \frac{E_{\text{axial}}}{2} \]
Substituting the values:

\( \implies E_{\text{equatorial}} = \frac{2.67 \times 10^5 \text{ N/C}}{2} \approx 1.33 \times 10^5 \text{ N/C} \default \)
The direction of this field is opposite to the direction of the dipole moment vector.
In simple words: At a distance of 15 cm, the electric field along the axis of the dipole is \( 2.67 \times 10^5 \text{ N/C} \). Along the side (equatorial plane), the field is exactly half of that strength, which is \( 1.33 \times 10^5 \text{ N/C} \).

Exam Tip: State the short-dipole condition (\( r \gg a \)) explicitly before using the simplified formulas to show that your approximation is physically justified.