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Electric Field Due To Point Charges

Learn Electric Field Due to Point Charge with examples. Calculation of Electric Field in 2-D. Force on Charge in an Electric Field and Component of Electric Field on Equatorial Line of Dipole.

Electric Field Due to Point Charge

  • Consider a charged particle  Q in space. A test charge q is placed at a distance r. Charge Q doesn't interact with q directly, rather it produce an electric field  in space around it. This field interact with q and exerts force on it.

  • The electric field has its own existence and is not affected by q.

Now, if Q is displaced from its position by a distance d, field doesn't propagate instantaneously. It takes finite time to propagate which is given as :

\(t=\dfrac {r}{c}\)

where,

\(c\) is the speed of light.

 

Illustration Questions

Consider an electric field due to charge Q at a point P. Now, if the charge Q is displaced to some other place, it's field at P will change

A Instantaneously  

B After time, t = \(\dfrac {r}{c}\)

C Can't be said

D

×

Charge Q produce its own electric field in space around it. So, when Q is displaced from its position, the field doesn't propagate instantaneously. It takes finite time to propagate. 

\(t=\dfrac{r}{c}\);

where, \(c\) is the speed of light.

Consider an electric field due to charge Q at a point P. Now, if the charge Q is displaced to some other place, it's field at P will change

image
A

Instantaneously

 

.

B

After time, t = \(\dfrac {r}{c}\)

C

Can't be said

D

Option B is Correct

Calculation of Electric Field due to Point Charge

Electric Field on q

  • Consider a charge q placed near point charge Q.

  • Force on q due to Q

\(\vec F =\dfrac {k\,Q\,q}{r^2}\hat r\)

  • Electric field at P is defined as

\(\vec E = \dfrac {\vec F}{q}=\dfrac {kQ}{r^2}\hat r\)

  • Electric field is a vector quantity.

NOTE:  Test charge q also creates an electric field. But charges don't exert forces on themselves. So, q is only measuring field of charge Q.

Effect of q on charge Q

  • Charge 'q' also exerts force on Q. So, it can disturb the position of Q and hence its field.
  • To ensure q, doesn't disturb other charge-
  1. The magnitude of q should be kept small.

                                               \(\vec E\,=\,\dfrac{\,\,\vec F}{\lim\limits_{q\to 0} q}\)

      2.  Other way is to fix charge Q, then q can be of any magnitude.

NOTE: Electric field at any point in space due to charge Q is independent of magnitude of test charge q.

Illustration Questions

A point charge of magnitude Q = 5 C is kept at a point in space. Calculate the electric field at a distance r = 30 m from it.

A 5 × 107 N/C

B 5 × 105 N/C

C 2 × 107 N/C

D 1 × 107 N/C

×

Electric field at distance r due to charge Q is given as,

\(\vec E =\dfrac {k\,|Q|\,}{r^2}\)

where,  Q = charge

             r = distance of point from charge

image

\(\vec E=\dfrac {9\times10^{9}\times|5|}{30\times30}\, N/C\)

\(\vec E=5\times10^{7}\, N/C\)

image

A point charge of magnitude Q = 5 C is kept at a point in space. Calculate the electric field at a distance r = 30 m from it.

A

5 × 107 N/C

.

B

5 × 105 N/C

C

2 × 107 N/C

D

1 × 107 N/C

Option A is Correct

Illustration Questions

Calculate net electric field at point P due to four charges Q1 = 5 C, Q2 = 5 C, Q3 = 2 C, Q4 = 2 C.

A \(\vec E_{net}=\dfrac {7\;k\;r}{(r^2+a^2)^{3/2}}\hat i\)

B \(\vec E_{net}=\dfrac {14\;k\;r}{(r^2+a^2)^{3/2}}\hat i\)

C \(\vec E_{net}=\dfrac {2\;k\; r}{(r^2+a^2)^{3/2}}\hat i\)

D \(\vec E_{net}=\dfrac {k\;r}{(r^2+a^2)^{3/2}}\;\hat i\)

×

  Figure representing of position of charges in 2-D plane i.e. Y-Z plane is as shown.

image

Position vector of each charge with respect to P

\(\vec r_{P1}= \vec r_P -\vec r_1=r\hat i-a\hat j\)

\(\vec r_{P2}= \vec r_P -\vec r_2=r\hat i+a\hat j\)

\(\vec r_{P3}= \vec r_P -\vec r_3=r\hat i-a\hat k\)

\(\vec r_{P4}= \vec r_P -\vec r_4=r\hat i+a\hat k\)

image

Electric field at P due to charge 1

\(\vec E_1=\dfrac {k|Q|}{\vec r_{P{1}}^{2}} \hat r_{P1} \Rightarrow\dfrac {k(5)}{(r^2+a^2)}\times \dfrac {(r\hat i-a\hat j)}{\sqrt{r^2+a^2}}\) (due to charge 1)

Similarly,

\(\vec E_2=\dfrac {k|Q|}{\vec r_{P{2}}^{2}} \hat r_{P2} \Rightarrow\dfrac {k(5)}{(r^2+a^2)}\times \dfrac {(r\hat i+a\hat j)}{\sqrt{r^2+a^2}}\) (due to charge 2)

\(\vec E_3=\dfrac {k|Q|}{\vec r_{P{3}}^{2}} \hat r_{P3} \Rightarrow\dfrac {k(2)}{(r^2+a^2)}\times \dfrac {(r\hat i-a\hat k)}{\sqrt{r^2+a^2}}\)(due to charge 3)

\(\vec E_4=\dfrac {k|Q|}{\vec r_{P{4}}^{2}} \hat r_{P4} \Rightarrow\dfrac {k(2)}{(r^2+a^2)}\times \dfrac {(r\hat i+a\hat k)}{\sqrt{r^2+a^2}}\)(due to charge 4)

image

Net electric field at P,

\(\vec E_{net}=\vec E_1+\vec E_2+\vec E_3+\vec E_4\)

\(=\dfrac {k(14)}{[(r^2+a^2)]^{3/2}} r\;\hat i\)

\(\vec E_{net}=\dfrac {k(14)\;r}{[r^2+a^2]^{3/2}}\hat i\)

image

From this, it is clear that if charges in Y-Z plane are symmetrically distributed about X-axis, then only x component of electric field gets added and others get canceled.

image

Calculate net electric field at point P due to four charges Q1 = 5 C, Q2 = 5 C, Q3 = 2 C, Q4 = 2 C.

image
A

\(\vec E_{net}=\dfrac {7\;k\;r}{(r^2+a^2)^{3/2}}\hat i\)

.

B

\(\vec E_{net}=\dfrac {14\;k\;r}{(r^2+a^2)^{3/2}}\hat i\)

C

\(\vec E_{net}=\dfrac {2\;k\; r}{(r^2+a^2)^{3/2}}\hat i\)

D

\(\vec E_{net}=\dfrac {k\;r}{(r^2+a^2)^{3/2}}\;\hat i\)

Option B is Correct

Calculation of Electric Field in 2-D

  • Consider a point charge q at (0, y). Electric field at point P(x, 0) will be:

\(\vec E = \dfrac {kq}{(x^2+y^2)}\;\hat r\)

where,

\(\hat r=\dfrac {x}{\sqrt{x^2+y^2}}\hat i- \dfrac {y}{\sqrt{x^2+y^2}}\hat j\)

\(\vec E = E_x\;\hat i +E_y\;\hat j\)

\(E_x=\dfrac {k\;q\;x}{(\sqrt{x^2+y^2})^3}\)

\(E_y=\dfrac {-k\;q\;y}{(\sqrt{x^2+y^2})^3}\)

 

Illustration Questions

A point charge of q = 5 C is placed at (0, 3). Calculate \(\vec E\) and  \(E_x\)  at (4, 0).

A \(36\times10^7 (4\;\hat i -3\;\hat j)\,N/C\),  \(1.44 × 10^9\, N/C\)

B \(36\times10^7 (4\;\hat i +3\;\hat j)\, N/C\),  \(2.44 × 10^9\, N/C\)

C \(36\times10^7 (4\;\hat i +6\;\hat j)\,N/C\),  \(1.44 × 10^{–9} \,N/C\)

D \(36\times10^7 (4\;\hat i -3\;\hat j)\,N/C\), \( 1.44 × 10^8\, N/C\)

×

Electric field at point P due to charge Q is given as 

\(\vec E = \dfrac {k|Q|}{r^2}\;\hat r\)

 

 

image

Position vector of charge q with respect to P

\(\vec r_{P1}=\vec r_P-\vec r_1\)

\(=x\hat i-y\hat j\)

 

 

image

Electric field at point P is given as

\(\vec E =\dfrac {k\,q}{(x^2+y^2)^{3/2}} (x\;\hat i - y\;\hat j)\)

image

 

\(\vec E =\dfrac {9\times10^9\times5}{[(4)^2+(3)^2]^{3/2}} (4\;\hat i -3\;\hat j)\)

\(\vec E =36\times10^7 (4\;\hat i -3\;\hat j)\,N/C\) 

 

 

image

X-component of electric field

\(E_x =36\times10^7 \times 4\)

\(E_x=1.44\times10^{9} \,N/C\)

image

A point charge of q = 5 C is placed at (0, 3). Calculate \(\vec E\) and  \(E_x\)  at (4, 0).

A

\(36\times10^7 (4\;\hat i -3\;\hat j)\,N/C\),  \(1.44 × 10^9\, N/C\)

.

B

\(36\times10^7 (4\;\hat i +3\;\hat j)\, N/C\),  \(2.44 × 10^9\, N/C\)

C

\(36\times10^7 (4\;\hat i +6\;\hat j)\,N/C\),  \(1.44 × 10^{–9} \,N/C\)

D

\(36\times10^7 (4\;\hat i -3\;\hat j)\,N/C\)\( 1.44 × 10^8\, N/C\)

Option A is Correct

Force on Charge in an Electric Field

 

  • Electric field of Q at a point P is defined in terms of test charge as:

\(\vec E = \dfrac {\vec F} {q}\)

 

 

 

  • \(\vec E\) is independent of the magnitude of test charge.
  • Force on charge q placed in an electric field is given as \(\vec F =q\vec E\).

 

Illustration Questions

A charge of 5 C is placed in electric field \(\vec E = (5\hat i + 6\hat j)\, N/C.\) Calculate force on the charge.

A \(2\hat i+3\hat j \;N\)

B \(25\hat i-30\hat j \;N\)

C \(30\hat i+25\hat j \;N\)

D \(25\hat i+30\hat j \;N\)

×

\(\vec F = q\vec E\)

where,

F = Force on charge,

E = Electric field,

q = charge

 

 

\(\vec F = 5(5\hat i+6\hat j)\,N\)

\(= 25\hat i+30\hat j \;N\)

 

 

 

\(\vec F = 5(5\hat i+6\hat j)\,N\)

\(= 25\hat i+30\hat j \;N\)

 

 

A charge of 5 C is placed in electric field \(\vec E = (5\hat i + 6\hat j)\, N/C.\) Calculate force on the charge.

A

\(2\hat i+3\hat j \;N\)

.

B

\(25\hat i-30\hat j \;N\)

C

\(30\hat i+25\hat j \;N\)

D

\(25\hat i+30\hat j \;N\)

Option D is Correct

Electric Field on Hanging Charge

  • Consider on electric field as shown.
  • Force on charge q

?\(\vec F =q\vec E\)

 

On X-axis

\(T\;sin\theta=F\) ...(i)

On Y-axis

\(T\;cos\theta=mg\) ....(ii)

Dividing (i) by (ii) both equation

\(tan\theta=\dfrac {F}{mg}\) or \(tan\theta=\dfrac {|q\vec E|}{mg}\)

Illustration Questions

A point charge q = 5 C of m = 5 kg is suspended by  an insulating string in a uniform electric field \(\vec E\) = 5 N/C as shown in figure. Calculate the angle  \('\theta\,'\) made by the string with vertical.

A \(tan^{-1} \left( \dfrac {1}{3} \right)\)

B \(tan^{-1} \left( \dfrac {1}{2} \right)\)

C \(tan^{-1} \left( 2 \right)\)

D \(tan^{-1} \left( 3 \right)\)

×

On X-axis,

\(Tsin\,\theta= 5 × 5\, N\)

 \(T sin\theta= 25\, N\)

 

image

On Y-axis,

 \(T cos\theta= 5 × 10 \,N\)

\(T cos \theta= 50 \,N\)

image

\(tan\theta=\dfrac {25}{50}\)

\(\theta =tan^{-1} \left ( \dfrac {1}{2} \right)\)

image

A point charge q = 5 C of m = 5 kg is suspended by  an insulating string in a uniform electric field \(\vec E\) = 5 N/C as shown in figure. Calculate the angle  \('\theta\,'\) made by the string with vertical.

image
A

\(tan^{-1} \left( \dfrac {1}{3} \right)\)

.

B

\(tan^{-1} \left( \dfrac {1}{2} \right)\)

C

\(tan^{-1} \left( 2 \right)\)

D

\(tan^{-1} \left( 3 \right)\)

Option B is Correct

Illustration Questions

Two point charges attached by a string of length  \(\ell=30\,m\) are placed in an electric field as shown. Given q1 = –5mC, q2 = 5mC, Breaking tension T= 5000 N. Calculate the maximum value of \(\vec E\).  

A 1.050 × 106 N/C

B 2.157 × 106 N/C

C 1.159 × 106 N/C

D 3.21 × 106 N/C

×

Tmax = 5000 N

image

\(\vec F = \dfrac {kq_1q_2}{r^2}= \dfrac {9\times10^9\times (5\times10^{-3})^2} {30\times30}\)

\(\vec F = 250 \,N\)

image

Under equilibrium,

\(q\vec E=F + T\)

\(5 × 10^{–3} ×\vec E= 250 + 5000\)

\(\vec E = \dfrac {5250}{5\times10^{-3}}\)

\(\vec E= 1.050 × 10^6 \,N/C\)

image

Two point charges attached by a string of length  \(\ell=30\,m\) are placed in an electric field as shown. Given q1 = –5mC, q2 = 5mC, Breaking tension T= 5000 N. Calculate the maximum value of \(\vec E\).  

image
A

1.050 × 106 N/C

.

B

2.157 × 106 N/C

C

1.159 × 106 N/C

D

3.21 × 106 N/C

Option A is Correct

Component of Electric Field on 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 Electric Field on Equatorial Line

  • To calculate electric field at a point on equatorial line, consider a dipole and a point P on equatorial line of dipole at a distance r from its mid-point.

  • Using superposition principle, it can be concluded that net electric field at point P is the vector sum of an electric field at P due to -q and +q.

\(\vec E_{+q}=\dfrac {1}{4\pi\epsilon_0}\times \dfrac {q}{(\sqrt{r^2+\ell^2})^2}\) (B to P)

 and \(\vec E_{-q}=\dfrac {1}{4\pi\epsilon_0}\times \dfrac {q}{(\sqrt{r^2+\ell^2})^2}\) (P to A)

Here, \(|\;\vec E_{+q}\;|=|\;\vec E_{-q}\;|=E\)

  • At P

 \(\vec E_{net}=2E\,cos\theta\;(-\hat i)\)

\(\vec E_{net}=\dfrac {2\times1}{4\pi\epsilon_0}\times \dfrac {Q}{({r^2+\ell^2})}\,cos\theta\;(-\hat i)\)

 

 

 

From the triangle

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

\(\vec E_{net}=\dfrac {2}{4\pi\epsilon_0}\times \dfrac {Q}{({r^2+\ell^2})}\,\times \dfrac {\ell}{\sqrt{r^2+\ell^2}} \;(-\hat i)\)

\(\vec E_{net}=\dfrac {2\;Q\;\ell} {4\pi\epsilon_0\;{(r^2+\ell^2})^{3/2}} \;(-\hat i)\)

Illustration Questions

For the given system if, q1 = –2 \(\mu\)C, q2 = 2 \(\mu\)C, r = 4 mm and \(\ell\)= 6 mm, then electric field at P will be-

A 880  \((\hat i)\)V/m

B 86 × 107 \((-\hat i)\)V/m

C 98 \((\hat i)\) V/m

D 3 × 105 \((\hat j)\)V/m

×

\(E=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {2 \mu\,C }{(5mm)^2}=\dfrac {9\times10^9\times2\times10^{-6}} {25\times10^{-6}}\)

\(E=\dfrac {18}{25}\times10^9\, V/m\)

image

\(\vec E_{net}=2E\;cos\theta\;(-\hat i)\)

image image

For \(cos\theta\)

\(cos\theta=\dfrac{3}{5}\)

image image

\(\vec E_{net}=2 \times\dfrac {18}{25}\times10^9\times\dfrac {3}{5}(\;-\hat i)\)

\(\vec E_{net}=\dfrac {108}{125}\times10^9(\;-\hat i)\, V/m\)

\(\vec E_{net}=0.86\times10^9 (\;-\hat i)\, V/m\)

\(\vec E=86\times10^7(\;-\hat i\;)\, V/m\) 

image

For the given system if, q1 = –2 \(\mu\)C, q2 = 2 \(\mu\)C, r = 4 mm and \(\ell\)= 6 mm, then electric field at P will be-

image
A

880  \((\hat i)\)V/m

.

B

86 × 107 \((-\hat i)\)V/m

C

98 \((\hat i)\) V/m

D

3 × 105 \((\hat j)\)V/m

Option B is Correct

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