Potential-Due-to-Different-Charge-Distributions L-4 Potential due to Different Charge Distribution
→ \rightarrow → → \rightarrow → Potential due to Different Charge Distribution → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Electrostatics Potential Energy
Potential energy of a pair of point charges Q & q is U = Q q 4 π ϵ 0 r U=\frac{Qq}{4 \pi \epsilon_0 r} U = 4 π ϵ 0 r Qq
Potential energy of a system of charges
U = q 1 q 2 4 π ϵ 0 r 12 + q 1 q 3 4 π ϵ 0 r 13 + q 2 q 3 4 π ϵ 0 r 23 U=\frac{q_1q_2}{4 \pi \epsilon_0 r_{12}}+\frac{q_1q_3}{4 \pi \epsilon_0 r_{13}}+\frac{q_2q_3}{4 \pi \epsilon_0 r_{23}} U = 4 π ϵ 0 r 12 q 1 q 2 + 4 π ϵ 0 r 13 q 1 q 3 + 4 π ϵ 0 r 23 q 2 q 3
→ \rightarrow → → \rightarrow → Electrostatics Potential Energy → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Electrostatics Potential
Electrostatics potential for a point charge Q.
V ( r ) = Q 4 π ϵ 0 r V(r)=\frac{Q}{4 \pi \epsilon_0 r} V ( r ) = 4 π ϵ 0 r Q
V(r) is a scalar quantity.
V ( r ) ⃗ V\vec{(r)} V ( r ) E ⃗ ( r ) ⃗ \vec{E}\vec{(r)} E ( r )
Unit of potential 1 Volt = 1 J/C
Potential due to Different Charge Distribution → \rightarrow → → \rightarrow → Electrostatics Potential → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Work Done
Unit of energy : electron volt(eV)
1 e V = ( 1 e ) × 1 V = 1.6 × 10 − 19 J 1eV=(1e)\times 1V=1.6 \times 10^{-19}J 1 e V = ( 1 e ) × 1 V = 1.6 × 1 0 − 19 J
Work done by an external force in moving a unit charge from a point r i r_i r i r f r_f r f
W = V ( r f ) − V ( r i ) W=V(r_f)-V(r_i) W = V ( r f ) − V ( r i )
For a point charge
W = Q 4 π ϵ 0 ( 1 r f − 1 r i ) W=\frac{Q}{4 \pi \epsilon_0}(\frac{1}{r_f}-\frac{1}{r_i}) W = 4 π ϵ 0 Q ( r f 1 − r i 1 )
Electrostatics Potential Energy → \rightarrow → → \rightarrow → Work Done → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Superposition Principle
Superposition Principle
V ( r ) = q 1 4 π ϵ 0 r 1 + q 2 4 π ϵ 0 r 2 + q 3 4 π ϵ 0 r 3 V(r) = \frac{q_1}{4 \pi \epsilon_0 r_1}+\frac{q_2}{4 \pi \epsilon_0 r_2}+\frac{q_3}{4 \pi \epsilon_0 r_3} V ( r ) = 4 π ϵ 0 r 1 q 1 + 4 π ϵ 0 r 2 q 2 + 4 π ϵ 0 r 3 q 3
= ∑ q i 4 π ϵ 0 r i = \sum\frac{q_i}{4 \pi \epsilon_0 r_i} = ∑ 4 π ϵ 0 r i q i
Distribution of charges
V ( r ) = 1 4 π ϵ 0 ∫ d q r 1 V(r)=\frac{1}{4 \pi \epsilon_0}\int\frac{dq}{r^1} V ( r ) = 4 π ϵ 0 1 ∫ r 1 d q
Electrostatics Potential → \rightarrow → → \rightarrow → Superposition Principle → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential Due to Conducting Sphere
Potential of a charged conducting sphere-
E ⃗ = Q r ^ 4 π ϵ 0 r 2 , r > R \vec{E}=\frac{Q \hat{r}}{4 \pi \epsilon_0 r^2},r>R E = 4 π ϵ 0 r 2 Q r ^ , r > R
V(r) = ∫ ∞ r \int_\infty^r ∫ ∞ r F ⃗ e n c \vec{F}_{enc} F e n c d r ⃗ d\vec{r} d r
= − Q 4 π ϵ 0 = -\frac{Q}{4 \pi \epsilon_0} = − 4 π ϵ 0 Q
∫ ∞ r d r r 2 \int_\infty^r \frac{dr}{r^2} ∫ ∞ r r 2 d r
= Q 4 π ϵ 0 r = \frac{Q}{4 \pi \epsilon_0 r} = 4 π ϵ 0 r Q
V ( R ) = Q 4 π ϵ 0 R V(R) = \frac{Q}{4 \pi \epsilon_0 R} V ( R ) = 4 π ϵ 0 R Q
Work Done → \rightarrow → → \rightarrow → Potential Due to Conducting Sphere → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential Due to Conducting Sphere
Superposition Principle → \rightarrow → → \rightarrow → Potential Due to Conducting Sphere → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Example
Conducting sphere of radius R = 10cm = 0.1m
Q = 1 n C = 10 − 9 C Q = 1nC=10^{-9}C Q = 1 n C = 1 0 − 9 C
V = Q 4 π ϵ 0 R = 10 − 9 × 9 × 10 9 0.1 = 90 V V=\frac{Q}{4 \pi \epsilon_0 R}=\frac{10^{-9}\times9\times10^9}{0.1}=90V V = 4 π ϵ 0 R Q = 0.1 1 0 − 9 × 9 × 1 0 9 = 90 V
E = Q 4 π ϵ 0 R 2 = V R = 90 0.1 = 900 V / m E=\frac{Q}{4 \pi \epsilon_0 R^2}=\frac{V}{R}=\frac{90}{0.1}=900V/m E = 4 π ϵ 0 R 2 Q = R V = 0.1 90 = 900 V / m
Potential Due to Conducting Sphere → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Example
Potential Due to Conducting Sphere → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential Due to a Dipole
V ( P ) = q 4 π ϵ 0 r 1 − q 4 π ϵ 0 r 2 V(P)=\frac{q}{4 \pi \epsilon_0 r_1}-\frac{q}{4 \pi \epsilon_0 r_2} V ( P ) = 4 π ϵ 0 r 1 q − 4 π ϵ 0 r 2 q
= q 4 π ϵ 0 ( 1 r 1 − 1 r 2 ) =\frac{q}{4 \pi \epsilon_0}(\frac{1}{r_1}-\frac{1}{r_2}) = 4 π ϵ 0 q ( r 1 1 − r 2 1 )
r 1 2 = r 2 + a 2 − 2 a r cos θ r_1^2 = r^2+a^2-2ar\cos\theta r 1 2 = r 2 + a 2 − 2 a r cos θ
2 2 2 = r 2 + a 2 + 2 a r cos θ 2_2^2=r^2+a^2+2ar\cos\theta 2 2 2 = r 2 + a 2 + 2 a r cos θ
Example → \rightarrow → → \rightarrow → Potential Due to a Dipole → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential Due to a Dipole
r 1 2 = r 2 + a 2 − 2 a r cos θ r_1^2=r^2+a^2-2ar\cos\theta r 1 2 = r 2 + a 2 − 2 a r cos θ = r 2 ( 1 + a 2 r 2 − 2 a r cos θ ) =r^2(1+\frac{a^2}{r^2}-\frac{2a}{r}\cos\theta) = r 2 ( 1 + r 2 a 2 − r 2 a cos θ )
r 1 = r ( 1 + a 2 r 2 − 2 a r cos θ ) 1 / 2 r_1=r(1+\frac{a^2}{r^2}-\frac{2a}{r}\cos\theta)^{1/2} r 1 = r ( 1 + r 2 a 2 − r 2 a cos θ ) 1/2
1 r 1 = 1 r ( 1 + a 2 r 2 − 2 a r cos θ ) − 1 / 2 \frac{1}{r_1}=\frac{1}{r}(1+\frac{a^2}{r^2}-\frac{2a}{r}\cos\theta)^{-1/2} r 1 1 = r 1 ( 1 + r 2 a 2 − r 2 a cos θ ) − 1/2
r » a
1 r 1 ≃ 1 r ( 1 + a r cos θ ) \frac{1}{r_1}\simeq\frac{1}{r}(1+\frac{a}{r}\cos\theta) r 1 1 ≃ r 1 ( 1 + r a cos θ )
Negative terms of order a 2 r 2 \frac{a^2}{r^2} r 2 a 2
Example → \rightarrow → → \rightarrow → Potential Due to a Dipole → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential Due to a Dipole
r 2 2 = r 2 + a 2 + 2 a r cos θ r_2^2=r^2+a^2+2ar\cos\theta r 2 2 = r 2 + a 2 + 2 a r cos θ
1 r 2 ≃ 1 r ( 1 − a r cos θ ) \frac{1}{r_2}\simeq\frac{1}{r}(1-\frac{a}{r}\cos\theta) r 2 1 ≃ r 1 ( 1 − r a cos θ )
1 r 1 − 1 r 2 ≃ 2 a r 2 cos θ \frac{1}{r_1}-\frac{1}{r_2}\simeq\frac{2a}{r^2}\cos\theta r 1 1 − r 2 1 ≃ r 2 2 a cos θ
V ( P ) = q 4 π ϵ 0 . 2 a r 2 cos θ V(P)=\frac{q}{4 \pi \epsilon_0}.\frac{2a}{r^2}\cos\theta V ( P ) = 4 π ϵ 0 q . r 2 2 a cos θ
∣ p ∣ ⃗ = q .2 q \vec{|p|}=q.2q ∣ p ∣ = q .2 q
V ( P ) = ∣ p ∣ ⃗ cos θ 4 π ϵ 0 r 2 = p ⃗ . g ^ 4 π ϵ 0 r 2 V(P)=\frac{\vec{|p|}\cos\theta}{4 \pi \epsilon_0 r^2}=\frac{\vec{p}.\hat{g}}{4 \pi \epsilon_0 r^2} V ( P ) = 4 π ϵ 0 r 2 ∣ p ∣ c o s θ = 4 π ϵ 0 r 2 p . g ^
Potential Due to a Dipole → \rightarrow → → \rightarrow → Potential Due to a Dipole → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential Due to a Dipole
V ( r ) = p ⃗ . r ^ 4 π ϵ 0 r 2 V(r)=\frac{\vec{p}.\hat{r}}{4 \pi \epsilon_0 r^2} V ( r ) = 4 π ϵ 0 r 2 p . r ^
V ( r ) = P cos θ 4 π ϵ 0 r 2 V(r)=\frac{P\cos\theta}{4 \pi \epsilon_0 r^2} V ( r ) = 4 π ϵ 0 r 2 P c o s θ
θ = 0 ; V ( r ) = p 4 π ϵ 0 r 2 \theta=0;V(r)=\frac{p}{4 \pi \epsilon_0 r^2} θ = 0 ; V ( r ) = 4 π ϵ 0 r 2 p
θ = π ; V ( r ) = − p 4 π ϵ 0 r 2 \theta=\pi;V(r)=-\frac{p}{4 \pi \epsilon_0 r^2} θ = π ; V ( r ) = − 4 π ϵ 0 r 2 p
θ = π 2 ; V ( r ) = 0 \theta=\frac{\pi}{2};V(r)=0 θ = 2 π ; V ( r ) = 0
Potential Due to a Dipole → \rightarrow → → \rightarrow → Potential Due to a Dipole → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential of an Infinite Linear Charge Density
2 π r . l . E = λ l ϵ 0 2 \pi r.l.E=\frac{\lambda l}{\epsilon_0} 2 π r . l . E = ϵ 0 λ l
E ⃗ = λ 2 π ϵ 0 r . r ^ \vec{E}=\frac{\lambda}{2\pi\epsilon_0r}.\hat{r} E = 2 π ϵ 0 r λ . r ^
W = − i n t r a r b λ r ^ 2 π ϵ 0 r . r ^ d r W=-int_{r_a}^{r_b}\frac{\lambda \hat{r}}{2 \pi \epsilon_0 r}.\hat{r}dr W = − in t r a r b 2 π ϵ 0 r λ r ^ . r ^ d r
= − λ 2 π ϵ 0 ∫ r a r b d r r = λ 2 π ϵ 0 l n ( r a r b ) =-\frac{\lambda}{2 \pi \epsilon_0}\int_{r_a}^{r_b}\frac{dr}{r}=\frac{\lambda}{2 \pi \epsilon_0}ln(\frac{r_a}{r_b}) = − 2 π ϵ 0 λ ∫ r a r b r d r = 2 π ϵ 0 λ l n ( r b r a )
Potential Due to a Dipole → \rightarrow → → \rightarrow → Potential of an Infinite Linear Charge Density → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential of an Infinite Linear Charge Density
Zero of potential at same r value
Let V=0 at r = r a = R r=r_a=R r = r a = R r b = r r_b=r r b = r
V ( r ) = λ 2 π ϵ 0 l n ( R r ) V(r)=\frac{\lambda}{2 \pi \epsilon_0}ln(\frac{R}{r}) V ( r ) = 2 π ϵ 0 λ l n ( r R )
Potential Due to a Dipole → \rightarrow → → \rightarrow → Potential of an Infinite Linear Charge Density → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Equipotential Surfaces
V = Q 4 π ϵ 0 r V=\frac{Q}{4 \pi \epsilon_0 r} V = 4 π ϵ 0 r Q
r = r 1 ; V = V 1 = Q 4 π ϵ 0 r 1 r=r_1 ; V=V_1=\frac{Q}{4 \pi \epsilon_0 r_1} r = r 1 ; V = V 1 = 4 π ϵ 0 r 1 Q
r = r 2 ; v = v 2 = Q 4 π ϵ 0 r 2 r=r_2;v=v_2=\frac{Q}{4 \pi \epsilon_0 r_2} r = r 2 ; v = v 2 = 4 π ϵ 0 r 2 Q
r 2 > r 1 r_2>r_1 r 2 > r 1
What about if
v 2 > v 1 v_2>v_1 v 2 > v 1 v 1 > v 2 ? v_1>v_2? v 1 > v 2 ?
Potential of an Infinite Linear Charge Density → \rightarrow → → \rightarrow → Equipotential Surfaces → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4
E ⃗ = E 0 k ^ \vec{E}=E_0\hat{k} E = E 0 k ^
Potential of an Infinite Linear Charge Density → \rightarrow → → \rightarrow → Uniform Electric Field → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Equipotentials of Point Charge
Equipotential Surfaces → \rightarrow → → \rightarrow → Equipotentials of Point Charge → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Equipotentials of a Dipole
Uniform Electric Field → \rightarrow → → \rightarrow → Equipotentials of a Dipole → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Electric Field and Potential
Consider two adjacent equipotentials surfaces
Work done by the external force in moving a unit charge from point A to point B
d W = ( v 0 + d v ) − v 0 dW=(v_0+dv)-v_0 d W = ( v 0 + d v ) − v 0
Work done in also given by − E ⃗ . d l ⃗ -\vec{E}.\vec{dl} − E . d l = − E d l cos θ =-Edl\cos \theta = − E d l cos θ
Equipotentials of Point Charge → \rightarrow → → \rightarrow → Electric Field and Potential → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Electric Field and Potential
E d l cos θ = − d v Edl\cos\theta=-dv E d l cos θ = − d v
E cos θ = − d v d l E\cos\theta=-\frac{dv}{dl} E cos θ = − d l d v
E x = − d v d x E_x=-\frac{dv}{dx} E x = − d x d v
E y = − d v d y E_y=-\frac{dv}{dy} E y = − d y d v
E z = − d v d z E_z=-\frac{dv}{dz} E z = − d z d v
Equipotentials of a Dipole → \rightarrow → → \rightarrow → Electric Field and Potential → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Potential of a Pair Charge
V ( r ) = Q 4 π ϵ 0 r V(r)=\frac{Q}{4 \pi \epsilon_0 r} V ( r ) = 4 π ϵ 0 r Q
= Q 4 π ϵ 0 ( x 2 + y 2 + z 2 ) 1 / 2 =\frac{Q}{4 \pi \epsilon_0(x^2+y^2+z^2)^{1/2}} = 4 π ϵ 0 ( x 2 + y 2 + z 2 ) 1/2 Q
E x = − d v d x = − Q ( − 1 / 2 ) .2 x 4 π ϵ 0 ( x 2 + y 2 + z 2 ) 3 / 2 E_x=-\frac{dv}{dx}=-\frac{Q(-1/2).2x}{4 \pi \epsilon_0 (x^2+y^2+z^2)^{3/2}} E x = − d x d v = − 4 π ϵ 0 ( x 2 + y 2 + z 2 ) 3/2 Q ( − 1/2 ) .2 x
= Q 4 π ϵ 0 ( x 2 + y 2 + z 2 ) x x 2 + y 2 + z 2 =\frac{Q}{4 \pi \epsilon_0 (x^2+y^2+z^2)}\frac{x}{\sqrt{x^2+y^2+z^2}} = 4 π ϵ 0 ( x 2 + y 2 + z 2 ) Q x 2 + y 2 + z 2 x
= Q 4 π ϵ 0 r 2 ( x r ) =\frac{Q}{4 \pi \epsilon_0r^2}(\frac{x}{r}) = 4 π ϵ 0 r 2 Q ( r x )
Electric Field and Potential → \rightarrow → → \rightarrow → Potential of a Pair Charge → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Problem
Electric Field and Potential → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Potential-Due-to-Different-Charge-Distributions L-4 Thank You
Potential of a Pair Charge → \rightarrow → → \rightarrow → Thankyou → \rightarrow → → \rightarrow →
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Potential-Due-to-Different-Charge-Distributions L-4 Potential due to Different Charge Distribution $\rightarrow$ $\rightarrow$ Potential due to Different Charge Distribution $\rightarrow$ Electrostatics Potential Energy $\rightarrow$ Electrostatics Potential
