Informative line

### 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.

#### 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 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.

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

$$\vec E=5\times10^{7}\, N/C$$ ### 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

#### 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. Position vector of each charge with respect to P

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

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

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

$$\vec r_{P4}= \vec r_P -\vec r_4=r\hat i+a\hat k$$ 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) Net electric field at P,

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

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

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

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

Option B 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

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

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

#### 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

×

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

$$E=\dfrac {18}{25}\times10^9\, V/m$$ $$\vec E_{net}=2E\;cos\theta\;(-\hat i)$$  For $$cos\theta$$

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

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

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

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

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,

$$\hat r=\dfrac {x}{\sqrt{x^2+y^2}}\hat i- \dfrac {y}{\sqrt{x^2+y^2}}\hat j$$  $$\vec E = E_x\;\hat i +E_y\;\hat j$$

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

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

#### 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

$$\vec E = \dfrac {k|Q|}{r^2}\;\hat r$$ Position vector of charge q with respect to P

$$\vec r_{P1}=\vec r_P-\vec r_1$$

$$=x\hat i-y\hat j$$ Electric field at point P is given as

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

$$\vec E =36\times10^7 (4\;\hat i -3\;\hat j)\,N/C$$ X-component of electric field

$$E_x =36\times10^7 \times 4$$

$$E_x=1.44\times10^{9} \,N/C$$   ### 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 $$\vec F =q\vec E$$.

#### 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 $$tan\theta=\dfrac {|q\vec E|}{mg}$$

#### 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,

$$Tsin\,\theta= 5 × 5\, N$$

$$T sin\theta= 25\, N$$ On Y-axis,

$$T cos\theta= 5 × 10 \,N$$

$$T cos \theta= 50 \,N$$ $$tan\theta=\dfrac {25}{50}$$

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

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

Option B is Correct

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

$$\vec F = 250 \,N$$ Under equilibrium,

$$q\vec E=F + T$$

$$5 × 10^{–3} ×\vec E= 250 + 5000$$

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

$$\vec E= 1.050 × 10^6 \,N/C$$ ### 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

Option A is Correct