Informative line

Coulombs Law

Learn Coulomb's law gives the electrostatic interaction force between the two point charges. Practice coulomb's law problems and representation of coulomb's law in vector form.

Coulomb's Law

  • Coulomb's law gives the electrostatic interaction force between the two point charges.
  • Electrostatic force between two point  charges separated by a distance r is given by Coulomb's law as 

         \(F_{12} = F_{21} = \dfrac{K|q_1||q_2|}{r^2}\)

where \(K\)= electrostatic constant = \(\dfrac{1}{4\pi\epsilon_0}=9× 10^9\,Nm^2/C^2\)

\(\epsilon_o = \) Permittivity of free space = \(8.85 × 10^{-12} C^2/ Nm^2\)

  • The electrostatic forces are  directed along the line joining the two point charges.
  • Repulsive force acts between two similar charges and attractive  force acts between opposite charges .

Note: Coulomb's law is only applicable for charges at rest.

Illustration Questions

Consider two point charges q1 and q2 separated by a distance r  in space. If the separation between them is doubled, then the magnitude of electric force between them is  

A Doubled 

B Halved 

C One - fourth

D Tripled 

×

Electric force between two point charges separated by distance r is given as 

\( F = \dfrac{K|q_1||q_2|}{r^2}\)

where  q1 , q2 = point charges 

r = separation between two point charges 

image

When the separation between point charges is doubled ,i.e.

\(r' = 2r\)

then electric force ,

\( F' = \dfrac{K|q_1||q_2|}{(r')^2} = \dfrac{K|q_1||q_2|}{(2r)^2}\)

\(F' = \dfrac{K|q_1||q_2|}{4\,r^2}\)

\(F' = \dfrac{1}{4}\left(\dfrac{K|q_1||q_2|}{r^2}\right)\)

\(F' = \dfrac{F}{4}\)

 

image

Hence, the magnitude of electric force becomes one - fourth.

image

Consider two point charges q1 and q2 separated by a distance r  in space. If the separation between them is doubled, then the magnitude of electric force between them is  

A

Doubled 

.

B

Halved 

C

One - fourth

D

Tripled 

Option C is Correct

Force on the Charge When Three Charges are Placed Along Straight Line 

  • Consider three point charges, q1 ,q2 and q3, placed along a straight line, as shown in figure.

  • To calculate net force on any one of the charge by the remaining two charges, we need to calculate the force on that charge by the other two charges and then add them with proper sign.
  • Suppose we want to calculate net force on charge q2.
  • Both the charges q1 and q3  exert  force on charge q2, as shown in figure.

  • Force on charge q2 due to q1 \((\vec F_{21})\) 

           Since charge q1 repels q2, Force \(\vec F_{21}\) acts towards right

             \(\vec F _{21} = \dfrac{K|q_2||q_1|}{(r_1)^2}\)   towards right

  • Force on charge q2 due to q3 \((\vec F_{23})\) 

         Since charge q3 repels q, Force \(\vec F_{23}\) acts towards left  

          \(\vec F _{23} = \dfrac{K|q_2||q_3|}{(r_2)^2}\)   towards left

  • Thus net force on charge q2

           \(\vec F_2 =\vec F_{21} +\vec F_{23} \)

           \(\vec F_2 = \dfrac{K q_1q_2}{r_1^2}\)  towards right  + \( \dfrac{K q_2q_3}{r_2^2}\)  towards left

  • Similarly net force on any charge can be calculated.

Illustration Questions

Consider three charges \(q_1 = 2\,\mu C\) ,  \(q_2 = 5\,\mu C\) and \(q_3 = -2\,\mu C\) placed along a straight  line as shown in figure. Charges q1 and q2 are placed  a = 1 m apart  and charges q2 and q3  are placed b= 3 m apart. Calculate net force on charge q2.

A 109 N towards right 

B 1010 N towards left 

C 0.1 N  towards right 

D 1012 N towards left 

×

Net force on charge \(q_2= 5\ C\)  is the sum of force  due to other two charges on it.

\(\vec F_2 =\vec F_{21} +\vec F_{23} \)

Force on charge q2 due to charge q1 \((\vec F _{21})\)

Since, charge q1 repels q2 , so \(\vec F _{21}\) acts towards right 

\(\vec F_{21} = \dfrac{K |q_1||q_2|}{a^2}\)  towards right 

\(\vec F_{21} = \dfrac{9× 10^9 × 2× 10^{-6}× 5× 10^{-6} }{(1)^2}\)

\(\vec F_{21} = 9 × 10^{-2} \,N\)   towards right 

image

Force on charge q2 due to charge q3 \((\vec F _{23})\)

Charge q3 attracts charge q, so \(\vec F _{23}\) acts towards right 

\(\vec F_{23} = \dfrac{K |q_2||q_3|}{b^2}\)  towards right 

\(\vec F_{23} = \dfrac{9× 10^9 × 5× 10^{-6}×2× 10^{-6} }{(3)^2}\)

\(\vec F_{23} = 10^{-2} \,N\)   towards right 

image

Net force on charge \(q_2 =5C\) 

\(\vec F _2 = \vec F _{21} +\vec F _{23}\)

\(\vec F _2 = 9× 10^{-2}\) towards right +  \(10 ^{-2} \,\) towards right 

\(\vec F _2 = 10× 10^{-2}\)towards right

\(\vec F _2 =0 .1 \,N\)  towards right

Consider three charges \(q_1 = 2\,\mu C\) ,  \(q_2 = 5\,\mu C\) and \(q_3 = -2\,\mu C\) placed along a straight  line as shown in figure. Charges q1 and q2 are placed  a = 1 m apart  and charges q2 and q3  are placed b= 3 m apart. Calculate net force on charge q2.

image
A

109 N towards right 

.

B

1010 N towards left 

C

0.1 N  towards right 

D

1012 N towards left 

Option C is Correct

Representation of Coulomb's Law in Vector Form 

  • Consider two point charges q1 and q2 placed at 'r' distance apart, as shown in figure.

  • According to Coulomb's law, the electrostatic force between two point charges is given as 

              \(|\vec F_{12}| = |\vec F_{21}|= \dfrac{K |q_1||q_2|}{ r^2} \)

               here, \(|q_1|,|q_2|\) are magnitude of  both the charges

  • Thus, the above equation only gives the magnitude of force but not its direction.
  • Since, force is a vector quantity, so it is better to write Coulomb's law in vector form for its representation.

Case 1

 Consider two point charges, q1 and q2, placed at distance 'r' apart. 

Force on charge q1 due to q2 \((\vec F_{12})\)   in vector form 

  • Since, charges q1 and qare of same nature, so charge q2 repels q1 and force  \(\vec F_{12}\) acts away from q.

           \(\vec F_{12} = \dfrac{K q_1q_2}{|\vec r_{12}|^2} \hat r_{12}\)

here, qand q2  are the magnitude of charge along with sign

 \(\vec r_{12}\) is the position  vector of \(q_1\) with respect to  \(q_2 = \vec r_1 - \vec r_2 \)

where,  \(\vec r_1=\)position vector of q1

         \(\vec r_2=\)position vector of q2

\(\hat r_{12}\) is the unit vector directed  towards  q1 from q2

Conclusion - Hence, it is concluded that the direction of force \(\vec F_{12}\) is along \(\hat r_{12}\).

 Force on charge q2 due to q1 \((\vec F_{21})\) in vector form

  • Since charge are of same nature so, charge  qrepels q2 and force \(\vec F_{21}\)  acts away from q1 .

      \(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} \hat r_{21}\)

here, q1 and q2 are the magnitude of charge along with sign

 \(\vec r_{21}\) is the position  vector  of  \(q_2\) with respect to  \(q_1 = \vec r_2 - \vec r_1 \)

where,  \(\vec r_1=\)position vector of q1

           \(\vec r_2=\)position vector of q2

\(\hat r_{21}\) is the unit vector directed  towards  q2 from q1

Conclusion - Hence, it is concluded that the direction of force \(\vec F_{21}\) is along \(\hat r_{21}\).

Case 2

Consider two charges, q1 and – q2 , placed at 'r' distance apart.

Force on charge q1 due to – q2 \((\vec F_{12})\) in vector form 

  • Since q1 and qare opposite in nature so, –q2 attracts q1 and force \((\vec F_{12})\) will be towards –q2.  

\(\vec F_{12} = \dfrac{K q_1(-q_2)}{|\vec r_{12}|^2} \hat r_{12}\)

\(\vec F_{12} = \dfrac{K q_1q_2}{|\vec r_{12}|^2} (-\hat r_{12})\)

Conclusion -  From above formula it is proved that direction of force  is along \(- \hat r_{12}\).

  • Force on charge  –q2 due to q1 \((\vec F_{21})\) in vector form 

     Since, q1 and q2 are opposite in nature so, q1 will attract – q2 . So, force \((\vec F_{21})\) will act towards q1.

\(\vec F_{21} = \dfrac{K q_1(-q_2)}{|\vec r_{21}|^2} \hat r_{21}\)

\(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} (-\hat r_{21})\)

Conclusion - From above formula it is concluded that direction of force is along \(- \hat r_{21}\) .

 

Case 3

 Consider two charges, –q1 and –q2 ,  placed at 'r' distance apart.

  • Force on charge –q1 due to – q2 \((\vec F_{12})\) in vector form 

Since charge are of same nature so, –q2 repels –q1 and force \(\vec F_{12}\) acts away form  –q2.  

\(\vec F_{12} = \dfrac{K (-q_1)(-q_2)}{|\vec r_{12}|^2} \hat r_{12}\)

\(\vec F_{12} = \dfrac{K q_1q_2}{|\vec r_{12}|^2} \hat r_{12}\)

Conclusion -  Hence, It is concluded that the direction of force \(\vec F_{12}\) is along \( \hat r_{12}\) .

 

  • Force on charge  –q2 due to –q1  in vector form 

     Since, charge are same nature so, –q1 will repel – q2  and force \(\vec F_{21} \) will act away form - q1.

\(\vec F_{21} = \dfrac{K (-q_1)(-q_2)}{|\vec r_{21}|^2} \hat r_{21}\)

\(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} \hat r_{21}\)

Conclusion - Hence , it is concluded that direction of force \(\vec F_{21}\) is along \(\hat r_{21}\) .

Note - In vector form of Coulomb's law we don't need to care about the direction of force, just put the values of q1 and q2 along with sign, vector will take care of direction .

 

 

 

 

Illustration Questions

Consider two charges, \(q_1 = 2\ C\) and \(q_2 = – 3\ C\), separated by a distance \(r\) in space. Position vector of \(q_1\) with respect to \(q_2\) is \(\vec r_{12}\)  and that of \(q_2\)with respect to \(q_1\) is \(\vec r_{21}\).  Choose the correct  option regarding  the direction of force on charge  \(q_2\) due to \(q_1\).

A along \(\hat r_{21}\)

B along \(-\hat r_{21}\)

C along \(-\hat r_{12}\)

D along \(\hat r_{12}\)

×

Since, charges q1 and q2 are of opposite nature so, q1 attracts q2 and \(\vec F_{21}\) acts towards q1

Here ,  \(\vec F_{21}\) = force on q2  due  to q1 

\(\hat r_{21} =\) unit vector directed  towards  q2 from q1 

 

image

Force on charge q2 due to q1 

\(\vec F_{21} = \dfrac{K (q_1)(q_2)}{|\vec r_{21}|^2} \hat r_{21}\)

\(\vec F_{21} = \dfrac{K (2)(-3)}{|\vec r_{21}|^2} \hat r_{21}\)

\(\vec F_{21} = \dfrac{K (2)(3)}{|\vec r_{21}|^2}(- \hat r_{21})\)

So, force \(\vec F _{21}\) will be along \(-\hat r _{21}\) .

image

Consider two charges, \(q_1 = 2\ C\) and \(q_2 = – 3\ C\), separated by a distance \(r\) in space. Position vector of \(q_1\) with respect to \(q_2\) is \(\vec r_{12}\)  and that of \(q_2\)with respect to \(q_1\) is \(\vec r_{21}\).  Choose the correct  option regarding  the direction of force on charge  \(q_2\) due to \(q_1\).

image
A

along \(\hat r_{21}\)

.

B

along \(-\hat r_{21}\)

C

along \(-\hat r_{12}\)

D

along \(\hat r_{12}\)

Option B is Correct

Force on a Charge due to another Charge in Vector form

  •  Consider two point charges q1 and q2 placed at 'r' distance apart. 

Force on a charge q1 due to q2 \((\vec F_{12})\)  in vector form 

  • Since, charges q1 and qare of same nature, so charge q2 repels q1 and force  \(\vec F_{12}\) acts away form q2.

           \(\vec F_{12} = \dfrac{K q_1q_2}{|\vec r_{12}|^2} \hat r_{12}\)

here , qand q2  are the magnitude  of charge along with sign

 \(\vec r_{12}\) is the position  vector  of  \(q_1\) with respect to \(q_2 = \vec r_1 - \vec r_2 \)

where   \(\vec r_1=\)position vector of q1

            \(\vec r_2=\)position vector of q2

\(\hat r_{12}\) is the unit vector directed  towards  q1 from q2

Conclusion -  Hence, it is concluded that the direction of force \(\vec F_{12}\)   is along \(\hat r_{12}\).

  Force on charge q2 due to q1 \((\vec F_{21})\) is represented  in vector form  

  • Since, charges are of same nature  so, charge  qrepels q2 and force \(\vec F_{21}\)  acts away from q1

      \(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} \hat r_{21}\)

here, q1 and q2  are the magnitude of charge along with sign

 \(\hat r_{21}\) is the position  vector  of  \(q_2\) with respect to \(q_1 = \vec r_2 - \vec r_1 \)

where  \(\vec r_1=\)position vector of q1

          \(\vec r_2=\)position vector of q2

\(\hat r_{21}\) is the unit vector directed  towards  q2 from q1

Conclusion - Hence, it is concluded that the direction of force \(\vec F_{21}\) is along \(\hat r_{21}\).

Illustration Questions

There are two point charges  \(q_1 = - 3\,\mu C \) and \(q_2 = 3\,\mu C \) in space. Position vector of \(q_1\) is \(\vec r_1 = (2\hat i + 3\hat j)\ m\) and position  vector of \(q_2\) is \(\vec r_2 = (2\hat i + 6\hat j)\ m\). Calculate force on \(q_2\) due to \(q_1\). 

A \(9× 10^{–3} \,N (-\hat j)\)

B \(9× 10^6 \,N (-\hat i)\)

C \(9× 10^4\, N (\hat j)\)

D \(18× 10^9\, N (-\hat j)\)

×

Force on charge q2 due to q1 is given as 

\(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} \hat r_{21}\)

where  \(\vec r_{21}\) = position vector of q2 with respect to q

Position vector of q2 with respect to q1 ( \(\vec r_{21}\) )

\(\vec r_{21}\) =  position vector of q2 – position vector q1

\(\vec r_{21}=\vec r_2-\vec r_1\)

\(\vec r _{21} = (2 \hat i+6\hat j) - (2 \hat i+3\hat j)\)

\(\vec r_{21}=3\hat j\)

Direction of  \(\vec r_{21}=\hat r_{21}=\hat j\)  

Magnitude of  \(\vec r_{21}\)

     \(|\vec r _{21}| = 3\ m\) 

 

 

Given : \(q_1 = -3\ \mu C,\ \ \ q_2 = 3\ \mu C,\ \ \ |\vec r_{21}| = 3\ m,\ \ \ \hat r_{21} = \hat j\)           

\(\vec F_{21} = \dfrac{9× 10^9× (-3)(3) × 10^{-12}\hat j}{(3)^2}\)

\(\vec F_{21} = 9× 10^{-3}\, N (-\hat j)\)

There are two point charges  \(q_1 = - 3\,\mu C \) and \(q_2 = 3\,\mu C \) in space. Position vector of \(q_1\) is \(\vec r_1 = (2\hat i + 3\hat j)\ m\) and position  vector of \(q_2\) is \(\vec r_2 = (2\hat i + 6\hat j)\ m\). Calculate force on \(q_2\) due to \(q_1\). 

A

\(9× 10^{–3} \,N (-\hat j)\)

.

B

\(9× 10^6 \,N (-\hat i)\)

C

\(9× 10^4\, N (\hat j)\)

D

\(18× 10^9\, N (-\hat j)\)

Option A is Correct

Net Force on a Charge due to more Than Two Charges Using Coulomb's Law

  • Consider three charges, q1 ,q2 and q3 that are placed at the vertices of a triangle.

  • Position vector of charge \( q_1 = (x_1 \, \hat i + y_1\,\hat j)\) 

           Position vector of charge \( q_2 = (x_2 \, \hat i + y_2\,\hat j)\) 

           Position vector of charge \( q_3 = (x_3 \, \hat i + y_3\,\hat j)\) 

  • To calculate net force on charge q3, we need to calculate the force due to charge q1 and q2  on q3  individually and then add them.
  • Force due to charge  q1 on q

           \(\vec F_{31} = \dfrac{Kq_1q_3}{|\vec r_{31}|^2} \hat r_{31}\)

where position  vector \(\vec r_{31} = \vec r_3 -\vec r_1 \)

  • Force due to charge  q2 on q

\(\vec F_{32} = \dfrac{Kq_2q_3}{|\vec r_{32}|^2} \hat r_{32}\)

    where position  vector \(\vec r_{32} = \vec r_3 -\vec r_2 \)

  • Thus , net force on charge q3

\(\vec F _3 = \vec F_{31} + \vec F_{32}\)

\(\vec F_{3} = \dfrac{Kq_1q_3}{|\vec r_{31}|^2} \hat r_{31} + \dfrac{Kq_2q_3}{|\vec r_{32}|^2} \hat r_{32}\)

 

Illustration Questions

Three charge particles q1 , q2 and q3 are placed at the vertices of a triangle. Calculate the net force on charge q1.

A \(504 (-\hat i - \hat j)\ N\)

B \(270 (-\hat i + \hat j)\ N\)

C \(140 (-\hat i - \hat j)\ N\)

D \(120 (\hat i +\hat j)\ N\)

×

Position vector of q1 \((\vec r_1) = 0\hat i +0\hat j =0\)

Position vector of q2 \((\vec r_2) = 3\hat i +4\hat j \)

Position vector of q3 \((\vec r_3) = 4\hat i +3\hat j \)

 

Force on charge q1 due to q2 \((\vec F _{12})\)

           \(\vec F_{12} = \dfrac{Kq_1q_2}{|\vec r_{12}|^2} \hat r_{12}\)

Position vector of \(q_1\) with respect to \(q_2\),

            \(\vec r_{12} = \vec r_{1} -\vec r_{2} = (0\hat i + 0\hat j) - (3\hat i + 4\hat j)\)

                  \(=(-3\hat i - 4\hat j)\)

Magnitude of \(\vec r_{12} = \sqrt {(-3)^2+(-4)^2}\)

              \(=5\ m\)

Unit vector \(\hat r _{12} = \dfrac{\vec r_{12}}{|\vec r_{12}|} = \dfrac{-3 \hat i - 4\hat j}{5}\)

Force \(\vec F_{12} = \dfrac{9× 10^9 × 1× 10^{-3}× 1× 10^{-3}}{(5)^2} × \dfrac{(-3 \hat i - 4\hat j)}{5}\)

                 \(\vec F_{12}=\dfrac{9000}{125}(-3 \hat i - 4\hat j)\)

              \( \vec F_{12}=72(-3 \hat i - 4\hat j)\,N\)

Force on charge q1 due to q3 \((\vec F _{13})\)

               \(\vec F_{13} = \dfrac{Kq_1q_3}{|\vec r_{13}|^2} \hat r_{13}\)

Position vector of \(q_1\) with respect to \(q_3\)

            \(\vec r_{13} = \vec r_{1} -\vec r_{3} = 0 - (4\hat i + 3\hat j) = (-4\hat i - 3\hat j) \)

           Magnitude of \(\vec r_{13} = \sqrt {(-4)^2+(-3)^2}\)

              \(=5\ m\)

Unit vector \(\hat r _{13} = \dfrac{\vec r_{13}}{|\vec r_{13}|} = \dfrac{-4 \hat i - 3\hat j}{5}\)

Force \(\vec F_{13} = \dfrac{9× 10^9 × 1× 10^{-3}× 1× 10^{-3}}{(5)^2} × \dfrac{(-4 \hat i - 3\hat j)}{5}\)

                 \(\vec F_{13}=\dfrac{9000}{125}(-4 \hat i - 3\hat j)\)

              \( \vec F_{13}=72(-4 \hat i - 3\hat j)\,N\)

 

Thus, net force on q1,

\(\vec F_1 =\vec F_{12}+\vec F_{13} \)

\(= 72 (-3\hat i -4\hat j)+ 72 (-4\hat i -3\hat j)\)

\(= 72 (-7\hat i -7\hat j)\)

\(= 72 × 7(-\hat i -\hat j)\)

\(= 504(-\hat i -\hat j)\, N\)

Three charge particles q1 , q2 and q3 are placed at the vertices of a triangle. Calculate the net force on charge q1.

image
A

\(504 (-\hat i - \hat j)\ N\)

.

B

\(270 (-\hat i + \hat j)\ N\)

C

\(140 (-\hat i - \hat j)\ N\)

D

\(120 (\hat i +\hat j)\ N\)

Option A is Correct

Net Force  on Charge Placed at Equatorial Line of Dipole

Dipole 

A dipole is a combination of two equal and opposite charges, separated by a small distance d.

 

Representation of  Axial and Equatorial Lines of Dipole

 

Calculation of Force on Equatorial Line of Dipole 

To calculate force on a charge placed at a point on equatorial line of dipole, consider a dipole and a point P on equatorial line of dipole at a distance  r.

 

  • The net force on charge q0 is the vector sum of force due to – q and + q . 
  • Force on charge q0 by q 

         \(\vec F_{+q} = \dfrac{1}{4\pi\epsilon_o} \dfrac{q.q_0}{(\sqrt {r^2 +\ell^2)}^2} \)      (B to P)

         \(\vec F_{+q} = \dfrac{1}{4\pi\epsilon_o} \dfrac{q.q_0}{(r^2 +\ell^2)} \)      (B to P)

  • Force on q0 by – q , 

        \(\vec F_{-q} = \dfrac{1}{4\pi\epsilon_o} \dfrac{q.q_0}{(\sqrt {r^2 +\ell^2)}^2} \)        (P to A)

       \(\vec F_{-q} = \dfrac{1}{4\pi\epsilon_o} \dfrac{q.q_0}{(r^2 +\ell^2)} \)            (P to A)

Here , \(|\vec F_{+q}| =|\vec F_{-q}| = F \)

  • Thus , net force on charge q 

  • Thus, net force = \(2 F cos\theta\)

        Net force = \(2 × \dfrac{1}{4\pi\epsilon_o} × \dfrac{q.q_0}{(r^2+\ell^2)} cos\theta\)

    Here , \(cos \theta = \dfrac{\ell}{\sqrt{r^2+\ell^2}}\)

     Net force = \(2 × \dfrac{1}{4\pi\epsilon_o} × \dfrac{q.q_0}{(r^2+\ell^2)} . \dfrac{\ell}{\sqrt{r^2+\ell^2}}\)

     Net force = \( \dfrac{2}{4\pi\epsilon_o} × \dfrac{q.q_0\,\ell}{(r^2+\ell^2)^{3/2}}\)

  • Net force in vector form

 

\(\vec F_{net} = \dfrac{2}{4\pi\epsilon_o} × \dfrac{q.q_0\,\ell}{(r^2+\ell^2)^{3/2}} (-\hat i)\)

 

Illustration Questions

A point charge of \(q_0= -1\,\mu C\) is placed at a point P at a distance \(r=1\ mm\) from the mid point of  dipole, as shown in figure.The charge of dipole \(-q = -2\ \mu C\) and \(q= 2\ \mu C\) are separated by a distance \(\ell = 2\ mm\). Calculate the net force on the charge q0 .

A \(9\sqrt2 × 10^3\ \hat i\,N \)

B \(6× 10^8\ \hat i\,N\)

C \(4× 10^3\ \hat i\,N\)

D \(2× 10^6\ \hat i\,N\)

×

Net force on charge \(q_0\)

  \(\vec F _{net} = 2 \,F cos\theta \,\hat i = 2 \dfrac{K \,q.q_0}{r^2} cos\,\hat i\)

\(\vec F _{net} = 2 × \dfrac{1}{4\pi \epsilon_o}\dfrac{(2× 10^{-6})(1× 10^{-6})}{(\sqrt2× 10^{-3})^2} \,cos\theta\, \hat i\)

 

image image

For \(cos \,\theta\),

\(cos \,\theta = \dfrac{1}{\sqrt 2} \)

 

 

image image

Thus , Net force \(\vec F_{net} = \dfrac{2× 9× 10^9× 2× 10^{-12}}{2× 10^{-6}} × \dfrac{1}{\sqrt2} \hat i\)

\(\vec F _{net} = 9 \sqrt 2 × 10^3\ \hat i\, N \)

image

A point charge of \(q_0= -1\,\mu C\) is placed at a point P at a distance \(r=1\ mm\) from the mid point of  dipole, as shown in figure.The charge of dipole \(-q = -2\ \mu C\) and \(q= 2\ \mu C\) are separated by a distance \(\ell = 2\ mm\). Calculate the net force on the charge q0 .

image
A

\(9\sqrt2 × 10^3\ \hat i\,N \)

.

B

\(6× 10^8\ \hat i\,N\)

C

\(4× 10^3\ \hat i\,N\)

D

\(2× 10^6\ \hat i\,N\)

Option A is Correct

Practice Now