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 12.1 COULOMB’S LAW 

COULOMB’S LAW  

Statement:  

The electrostatic force between two point charges is directly proportional to the product of the magnitude of charges and inversely proportional to the square of distance between them. 

 

Mathematical Expression: If two point charges 

q1and 

q2

 are separated by a distance 

r

, then the electrostatic force ‘

F

’ between them is expressed as: 

F=kq1q2r2

  where k  is the constant of proportionality and is expressed as 

k=14πε0

 

where ε

0

 is the permittivity of free space and its value in SI unit is 

8.85×10−12C2N m2

. 

 

VECTOR FORM OF COULOMB'S LAW 

The electrostatic force 

F⃗ 

 between two point charges ‘

q1

’ to ‘

q2

’ is expressed as:  

F⃗ =14πε0   q1q2r2rˆ

 

Here 

rˆ

 is the unit vector directed from ‘

q1

’ to ‘

q2

’. 

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COULOMB FORCE IS A MUTUAL FORCE 

Coulomb’s force is a mutual force, it means that if a charge ‘

q1

’ exerts a force on charge ‘

q2

’, then ‘

q2

’ also exerts an equal and opposite force on ‘

q1

’.  If charge ‘

q1

’ exerts an electrostatic force ‘

F→21

’ due to charge ‘

q2

’ and ‘

q2

’ exerts electrical force ‘

F→12

’ on charge ‘

q1

’ and, then 

F→12=−F→21

 

Proof: If 

r ˆ21

 represents the direction of force from charge ‘

q1

’ to ‘

q2

’ and 

r ˆ12

 is the unit vector which represent the direction of force from charge ‘

q2

’ to ‘

q1

’, then 

F→12=14πε0q1q2r2r ˆ21            −−−−−  (2)

 

F→12=14πε0q1q2r2r ˆ12            −−−−−  (2)

 

As 

rˆ21=−

rˆ12

 , so the eq. (1) becomes 

F→21=14πε0q1q2r2(− r ˆ12)

 

F→21=−14πε0q1q2r2r ˆ12

 

By eq. (2) 

F→12=−F→21

 

This expression shows that Coulomb force is a mutual force.                             

 

EFFECT OF DIELECTRIC MEDIUM ON ELECTRICAL FORCE BETWEEN TWO POINT CHARGES 

If the dielectric medium having relative permittivity ‘

εr

’ is placed between two point charges, then the electrical force will reduced by 

εr

-times. The expression of coulomb’s force between two point charges, when the dielectric medium is placed between them, is expressed as: 

F =14πε0εrq1q2r2

 

DIELECTRIC (Definition):  

An insulator, placed between two point charges, is referred as dielectric. 

POINT CHARGES (Definition):  

The charges whose sizes are very small as compared to the distance between them are called point charges. 

Numerical Related to Article “12.1 COULOMB’s LAW” 

12.1 Compare magnitudes of electrical and gravitational forces exerted on an object (mass = 10.0 g, charge = 20.0 µC) by an identical object that is placed 10.0 cm from the first. 

(G = 6.67 ×10−11 Nm2kg−2)

 

Given Data:   Masses  m1=m2=m=10 g=0.01 kg , Charges  q1=q2=q=20 µC=20×10−6C ,  Distance  r=10 cm=0.1 m

To Determine:  FeFg=?

Calculations:FeFg=(kq1q2r2)(Gm1m2r2)=kq1q2Gm1m2=kq2Gm2=9×109×(20×10−6)26.67×10−11×(0.01)2=5.4×1014

 

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12.2 Calculate vectorially the net electrostatic force on q as shown in the figure.  

 

 

 

Given Data: Charges  q1=1 µC=1×10−6C ,  q2=−1 µC=−1×10−6C ,  q=4 µC=4×10−6C

To Determine: Total Force on  q   F=?

Calculations: From Fig.  tanθ=0.80.6⟹θ=tan−1(0.80.6)=53°   Force Exerted by Charge  q1  on  q :   F1=kqq1r2=9×109×4×10−6×1×10−6(1)2=36×10−3 N     Force Exerted by Charge  q2  on  q :   F2=kqq2r2=9×109×4×10−6×1×10−6(1)2=36×10−3 N     Now  Fx=F1x+F2x=F1cosθ+F2cosθ=36×10−3×cos53°+36×10−3×cos(2π−53°)=0.043 N   And  Fy=F1y+F2y=F1sinθ+F2sinθ=36×10−3×sin53°+36×10−3×sin(2π−53°)=0 N   Magnitude of Resultant Force  F=F2x+F2y−−−−−−−√=(0.043)2+(0)2−−−−−−−−−−−−√=0.043 N   Direction of Resultant Force  tanθ=FyFx⟹θ=tan−1(00.043)=0°  (Resultant is along x-axis)  Resultant Force  F⃗ =0.043 iˆ

Example12.1: Charges 

q1=100 μC

 and 

q2=50 μC

 are located in xy plane at positions 

r1=3.0 jˆ

 and 

r2=4.0 iˆ

 respectively, where the distances are measured in meters. Calculate the force on 

q2

. 

 

Given Data: First Charge  q1=100 μC=100×10−6 C , Second Charge  q2=50 μC=50×10−6 C   Position Vector of First Charge : r⃗ 1=3.0 jˆ , Position Vector of Second Charge :r⃗ 2=4.0 iˆ

To Determine: Force on   q2  due to  q1 :  F⃗ 21=?

Calculations:  Position Vector of   q2   relative to  q1:       r⃗ 21=r⃗ 2−r⃗ 1=4 iˆ−3 jˆ   Distance between charges [Math Processing Error]  By Coulomb’s Law: [Math Processing Error]  Magnitude of [Math Processing Error]   Direction of [Math Processing Error]

 

MCQs Related to the Article “12.1 COULOMB’S LAW” 

1. Coulomb’s law is only applicable for 

(a) Big charges 

(b) Small charges 

(c) Point charges 

(d) All charges 

 

2. If the distance between two point charges is doubled, the force between them will become: 

(a) Doubled  

(b) Half 

(c) Three Times 

(d) One forth 

3. The constant k in Coulomb’s Law depends upon  

(a) Nature of medium(b) System of units(c) Intensity of charge(d) Both a & b  

4. The value of permitivity of free space 

   εo
    is: 

(a) 

8.85×10−12C2Nm2

 

(b) 

8.85×10−12Nm2C2

 

(c) 

8.85×10−12NmC2

 

(d) 

8.85×10−12NC2m2

 

 

5. The value of coulomb’s constant 

   k
    is: 

(a) 

9×109C2Nm2

 

(b) 

9×109Nm2C2

 

(c) 

9×109NmC2

 

(d) 

9×109NC2m2

 

 

6. Unit relative permitivity is 

(a) 

C2Nm2

 

(b) 

Nm2C2

 

(c) 

NC2m2

 

(d) no unit 

 

7. Presence of dielectric always: 

(a) Increases the electrostatic force 

(b) Decreases the electrostatic force 

(c) Does not effect the electrostatic force 

(d) Doubles the electrostatic force 

 

8. The value of relative permitivity for all the dielectrics is always: 

(a) Less than unity 

(b) Greater than unity 

(c) Equal to unity 

(d) Zero 

 

9. Relative permitivity of air is: 

(a) 1.06 

(b) 1.006 

(c) 1.0006 

(d) 1.6 

 

10. The force between two similar unit charges placed one meter apart in air is: 

(a) Zero  

(b) One newton 

(c) 

9×109 N

 

(d) 

9×1019 N

 

 

11. If the magnitude of charges and distance between them is doubled, then the force will be: 

(a) Doubled 

(b) Halved 

(c) Remain same  

(d) On forth  

 

12. When an insulating medium is placed between two charges, the electrostatic force: 

(a) Increases 

(b) decreases 

(c) zero 

(d) Remain Same 

 

13. The electrostatic force between two charges is 42 N. If we place a dielectric of 

    εr=2.1
    , between the charges, then the force become equal to: 

(a) 42N 

(b) 88.2 N 

(c) 2 N 

(d) 20 N 

MCQ # 1: (c)	MCQ # 2: (d)	MCQ # 3: (d)	MCQ # 4: (a)	MCQ # 5: (b)	MCQ # 6: (d)	MCQ # 7: (b)
MCQ # 8: (b)	MCQ # 9: (c)	MCQ # 10: (c)	MCQ # 11: (c)	MCQ # 12: (b)	MCQ # 13: (d)

 

ARTICLE “12.1 COULOMB’S LAW” IN PAST PAPERS 

Short Questions: (2 Marks) 

1. State coulomb’s law. Express its mathematical form. 

2. What is the effect of medium between the charges upon coulomb’s force? 

3. Define electrostatics and electric force. 

Long Question (5 Marks) 

1. State Coulomb’s law for electrostatic force. Discuss its vector form and show that  

   F⃗ 12=−F⃗ 21