Electric Field of a Point Charge: Formula and Derivation

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An electric field is a region or space around a charged body within which its influence can be felt by other charged bodies. Mathematically, it can be expressed as electric force per unit charge. 

  • The direction of the electric field is radially outward from a positive charge and radially inward to a negative charge.
  • Electric charge is a property of subatomic particles that makes them experience a force when placed in a magnetic or electric field.
  • A point charge is a hypothetical charge whether positive or negative, whose electric field is found to be at a particular distance. 
  • Electric fields arise from time-variable electric currents and electric charges.
  • Magnetic and electric fields are both expressions of the electromagnetic field, one of the four basic interactions (also known as forces) of nature.

Key Terms: Coulomb's Law, Electric Field, Electrostatic Field, Electric Charge, Electromagnetic Field, Absolute permittivity

Also Read: Electric Flux


What is Electric Field?

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Electrical fields (sometimes referred to as E-fields) refer to the physical field that surrounds electrically charged particles and exert forces on all other charged particles within the field, either attracting or repelling them.

  • The electric field between two charges is similar to the gravitational field between two masses because both of them obey the inverse-square law with distance.
  • The electric field can be visualized with a set of imaginary lines known as the electric field lines or lines of force.
  • Electric field lines are originated from a positive charge and terminated on a negative charge.

Electric field lines

Electric field lines


What is Electric Charge?

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Electric charge is the property of a particle due to which it can apply force on another charged or uncharged particle.

  • It is a scalar quantity and the SI unit is Coulomb (C).
  • Dimensional formula of electric charge is [AT].
  • The charge associated with a proton is known as a positive charge.
  • The charge associated with an electron is known as a negative charge.

Electric Field of a Point Charge Formula

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The electric field intensity (E) due to a point charge (Q) at any point in its electric field is defined as the electrostatic or Coulomb’s force (F) per unit of charge exerted on an infinitesimal positive test charge (qo) at rest at that point.

Mathematically, the electric field intensity is given by

E = F/qo

  • It is a vector quantity.
  • SI unit of electric field intensity is Newton per coulomb (NC-1) or Volt per meter (Vm-1).
  • Dimensional formula of electric field intensity is [MLT-3A-1].

Electric field due to a point charge

Electric field due to a point charge

According to Coulomb’s law, the force between two charges Q and qo separated by distance r is given by

F = \(\frac{1}{4 \pi \epsilon_o} \frac{Qq_o}{r^2}\)

Where ∈o is the absolute permittivity of free space.

Therefore, the electric field intensity is given by

E = \(\frac{F}{q_o}\)\(\frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}\)

Above expression represents the electric field due to a point at distance r.

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Derivation of Electric Field Due to Two Point Charges

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Consider two point charges q1 and q2 placed at A and B having position vector \(\vec{r_1}\) and \(\vec{r_2}\)

The force on qo due to q1 is given by

F1 = \(\frac{1}{4 \pi \epsilon_0} \frac{q_0q_1}{|\vec{r}- \vec{r_1}|^3} (\vec{r} - \vec{r_1})\)

Force on qo due to q2 is given by

F2 = \(\frac{1}{4 \pi \epsilon_0} \frac{q_0q_2}{|\vec{r}- \vec{r_2}|^3} (\vec{r} - \vec{r_2})\)

Electric field due to two pint charges

Electric field due to two pint charges

Net force acting on qo placed at point P is given by

F = F1 + F2

⇒ F = \(\frac{1}{4 \pi \epsilon_0} \frac{q_0q_1}{|\vec{r}- \vec{r_1}|^3} (\vec{r} - \vec{r_1})\) + \(\frac{1}{4 \pi \epsilon_0} \frac{q_0q_2}{|\vec{r}- \vec{r_2}|^3} (\vec{r} - \vec{r_2})\)

⇒ F = \(\frac{q_0}{4 \pi \epsilon_0}\) \([\frac{q_1}{|\vec{r}- \vec{r_1}|^3} (\vec{r} - \vec{r_1}) + \frac{q_2}{|\vec{r}- \vec{r_1}|^3} (\vec{r} - \vec{r_2})]\)

Electric field intensity at point P, is given by

\(\vec{E}  = \frac{\vec{F}}{q_o}\)

\(\vec{E}\) = \(\frac{1}{4 \pi \epsilon_0}\) \([\frac{q_1}{|\vec{r}- \vec{r_1}|^3} (\vec{r} - \vec{r_1}) + \frac{q_2}{|\vec{r}- \vec{r_1}|^3} (\vec{r} - \vec{r_2})]\)


Electric Field Due to a System or Group of Point Charges

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The electric field intensity at any point because of a system or group of charges is equivalent to the vector sum of electric field intensities because of individual charges at the same point. Currently, we would perform the vector sum of electric field intensities:

\(\vec{E} = \vec{E_1} + \vec{E_2} + \vec{E_3} + … + \vec{E_n}\)

\(\vec{E_1}\) = \(\frac{1}{4 \pi \epsilon_0}\) \(\sum _{i = 1} ^ {i=n} \frac{Q_i}{r_i^2} (\hat{ri})\)

Here, 

  • ri is the distance of the point P from the ith charge Qi 
  • ri is a unit vector directed to the point P from Qi.

Let’s say charge Q1, Q2...Qn is positioned in a vacuum at places r1, r2...., and r? respectively.

Electric field due to multiple point charges

Electric field due to multiple point charges

The net forces at P are the vector sum of forces because of individual charges, given by,

\(\vec{F}\) = \(\frac{1}{4 \pi \epsilon_0} q_0 \sum_{i=1}^{i=n} \frac{Q_i}{|\vec{r}- \vec{ri}|^3 ×|\vec{r}- \vec{ri}|}\)

As, \(\vec{E}  = \frac{\vec{F}}{q_o}\)

Therefore,

\(\vec{E}\) = \(\frac{1}{4 \pi \epsilon_0} \sum_{i=1}^{i=n} \frac{Q_i}{|\vec{r}- \vec{ri}|^3 ×|\vec{r}- \vec{ri}|}\)

Putting \(\frac{1}{4 \pi \epsilon_0}\) = k

\(\vec{E}\) = \(k\frac{Q_1}{r_1^2}+ k\frac{Q_2}{r_2^2} +.............+ k\frac{Q_n}{r_n^2}\)


Electric Field due to Continous Distribution of Charge

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A system of closed spaced electric charges forms a continuous charge distribution. It can be on a line, on a surface, or on the entire volume.

To calculate the electric field due to charge distribution, the concept of charge density is introduced.

  1. Linear charge density (λ): It is defined as the charge per unit length of the conductor. It is given by

λ = q/l

  1. Surface charge density (σ): It is defined as the charge per unit area of the conductor. It is given by

σ = q/A

  1. Volume charge density (ρ): It is defined as the charge per unit volume of the conductor. It is given by

ρ = q/V

Electric field intensity due to Linear charge distribution is given by

E = \(\frac{F}{q_o}\) = \(\frac{1}{4 \pi \epsilon_o} \int _L \frac{ \lambda dL}{r^2}\)

Electric field intensity due to Surface charge distribution is given by

E = \(\frac{F}{q_o}\) = \(\frac{1}{4 \pi \epsilon_o} \int _A \frac{ \sigma dA}{r^2}\)

Electric field intensity due to Volume charge distribution is given by

E = \(\frac{F}{q_o}\) = \(\frac{1}{4 \pi \epsilon_o} \int _V \frac{ \rho dV}{r^2}\)

Also Read:


Things to Remember

  • An electric field is a physical field encompassing electrically charged particles and exercising a force on all other charged particles in the field, either attracting or repulsing them.
  • When an electric charge q₀ is positioned in close proximity to another charge Q, q₀ witnesses either an attractive or repulsive force.
  • Electric field due to a point charge formula is given by E = \(\frac{F}{q_o}\) = \(\frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}\)
  • Electric Field Due to two point charges is given by

\(\vec{E}\) = \(\frac{1}{4 \pi \epsilon_0}\) \([\frac{q_1}{|\vec{r}- \vec{r_1}|^3} (\vec{r} - \vec{r_1}) + \frac{q_2}{|\vec{r}- \vec{r_1}|^3} (\vec{r} - \vec{r_2})]\)

  • A system of closed spaced electric charges forms a continuous charge distribution. It can be on a line, on a surface, or on the entire volume.
  • Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is inversely proportional to the square of the distance between them and straight proportional to the magnitude of the charges’ product.

Sample Questions

Ques. Is electric charge a vector quantity? (2 Marks)

Ans. No, electric charge is a scalar quantity. In addition to magnitude and direction, a vector quantity is denoted by the rules of vector addition. Like the triangle law of vector addition and the parallelogram law of vector addition; only then the set is called a vector set.

When two currents encounter at a junction, as in the case of an electric current, the resulting current will be an algebraic sum and not a vector sum. Therefore, electric current is a scalar quantity, although it has magnitude and direction.

Ques. What do you mean by charging, and what are the various ways of charging? (3 Marks)

Ans. The technique of losing the electric charge from an object or providing the electric charge to an object is referred to as charging.

An uncharged object can be charged in three various methods as follows:

  • Charging by friction (triboelectric charging)
  • Charging by conduction
  • Charging by induction

Ques. If a point charge of 20μC is at a distance of 1m, calculate the magnitude of the point charge in the electric field. (2 Marks)

Ans. The formula for calculating the magnitude of the electric field due to a point charge will be:

E = \(\frac{kq}{d^2}\)

= \(\frac{9× 10^9 (20× 10^{-6})}{(1)^2}\)

= 1.8 × 105 N/C

Ques. Compute the magnitude of an electric field at a point (in the middle of two charges). These 2-point charges are 4 μC and – 3.2 μC, separated by a distance of 4 cm. (3 Marks)

Ans. Suppose the line that joins the charges is the x-axis. Thus, we will discover the electric field due to the charge at the midpoint (d = 2 cm). The electric field’s magnitude will be as mentioned below:

E1 = \(\frac{kq}{d^2}\)

E1 = 9 × 109(4 × 10-6) (0.02)2

E1 =90 × 106 N/C

Similarly, for – 3.2 μC

E2 = 9 × 109( – 3.2 × 10-6)(0.02)2

E2 = – 72 × 106 N/C

Thus, the resultant electric field at the midpoint will be

E = E2 + E1

= - 72 × 106 + 90× 106

 E = 18 × 106 N/C Magnitude

Therefore, in the middle of two charges, the magnitude of an electric field at a point will be 18 x 106 N/C. and with regard to – 3.2 μC, is 72 x 106 N/C.

Ques. Compute both the direction and magnitude of an electric field at a point of 2 cm if it is left to the point charge of – 2.4 nC. (3 Marks)

Ans. To calculate the magnitude of the electric field due to a point charge, we need only the absolute values (without any sign). So, we can just ignore the minus sign while calculating magnitude.

E = \(\frac{kq}{d^2}\)

= 9 × 109 (2.4 × 10-9)(0.02)2

= 54 × 103 N/C

Thus, the magnitude of this electric field is 54 × 103 N/C, and its direction is toward the charge.

Ques. A point charge q causes a test charge to be carried in an electric field from A to points B, C, D, and E lying on the same circle around q. Calculate the work done. (3 Marks)

Ans. Potential at A due to point charge = \(\frac{kq}{r}\) (radius of sphere = r)

Potential at B due to point charge = \(\frac{kq}{r}\)

The work completed for moving the test charge from A to B
= \(\frac{kq}{r}\)\(\frac{kq}{r}\)

Likewise for AD = AC = AE = AB = 0 (Work completed)

Ques. At what point along the axis is the electric field zero if you have two charges on one axis? A charge of −2 μC is placed at the origin, and another charge of −8 μC is placed at 4m. (3 Marks)

Ans. The electric field equation is from a point charge

E =\(\frac{kq}{r^2}\)

To find the point where the electric field is 0, we put the equations to have the same value for both charges since they will cancel each other from there. 

Let x be the locus of a point. The radius of the first charge is going to be x, while the radius for the second is going to be (4−x).

\(\frac{kq_1}{x^2}= \frac{kq_2}{(4-x)^2}\)

⇒ q1⋅(4−x)2 = q2x2

⇒ −2⋅(4−x)2 = −8x2

⇒ (4 - x)2 = 4x2

⇒ 4 – x = 2x

⇒ 4 = 3x

⇒ x = 4/3

Hence, the point at which the electric field is zero is at 4/3m or 1.34m.

Ques. An object of mass 25kg moves with an acceleration of 2.1 m/s2 in an electric field of 4.5 N/C. Calculate the charge of the object. (3 Marks)

Ans. Relate Newton's second law to the equation for the electric force due to an electric field:

Fnet = ma = FE = qE

Plug in values:

25 x 2.1 = 4.5q

q = 11.66C

Ques. Suppose two-point charges are divided by 5 meters. One includes a charge of 5 μC and the other includes a charge of – 2 μC. Calculate the magnitude of the force between them and decide whether is it attractive or repulsive. (3 Marks)

Ans. The equation for the force felt by two-point charges is referred to as Coulomb's law, and is as follows:

F = k q1q2 / r2

The value 'k' is referred to as Coulomb's constant, and its value is approximately 9.0 × 109 N. We have all the numbers we need to use this equation, so we can just plug it in.

= (9 × 109)(5 × 10-6)(-2 × 10-6)52

= − 0.0036

= – 3.6 × 10-3 N

Since we are given a negative number (and by our intuition: "opposites attract"), we can determine that the force is attractive. Since we are asked for the magnitude of the force, we take the absolute value, so our answer will be – 3.6 × 10-3 N, attractive force.

Ques. Calculate the value of the electric field 3 m from a point charge of strength 10 μC. (3 Marks)

Ans. To find the electric field strength produced by a point charge, you apply the below-mentioned equation.

E = k × Q/r2

Since we know the value of Q and r (charge and distance, respectively), we can simply plug in the numbers we have to discover the answer.

E = 9 x 109 [(10 x 10−6) / 32]

= 10000 N/C

This may seem like a very large number coming from such a small charge, remember that the typical charges interacting with it will be of roughly the same strength. This gives a force much less than 10,000 N.
 

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