Electric potential and potential difference and electric potential due to point charge
1st topic
“Electric potential”:
Electric potential is the physical quantity by which we can determine the direction of flow of charge. It is represented by the “V”.
*Measurment: Electric potential at a point in an electric field is equal to the work done in bringing a unit positive test charge from infinity to that point.
Electric potential, V = W/q•
Units :
1) SI unit = joule/coulomb or volt
2) CGS unit = erg/stat-coulomn or stat-volt
Dimensional
Formula = [ML^2T^-3A^-1]
*Relationship between volt and stat-volt ---
(Electric
potential )= work done/charge
1 volt = 1 joule/ 1coulomb
= 10^7 erg/ 3×10^9 stat-coulomb
1 volt = 1/300 stat-volt
2nd topic
*“Potential difference”: It is the work done in bringing a unit positive test charge from one point to another point. It is represented by V or ΔV.
It is scalar quantity.
According to definition ----
V2 - V1 = W/ q•
ΔV = (F. S. CosΦ) / q•
[ here Φ = 180 and F = q. E ]
ΔV = [(q• E)ΔX. CosΦ]/ q•
ΔV = E. ΔX. (-1)
ΔV = - E. Δx
E = - ΔV/ ΔX
[ E = - dv/ dx ]
Electric field intensity = negative potential gradient
3rd topic:
*“Electric potential due to point charge”:
In above figure a point charge +q be placed at point 0. We are to find the electric potential due to this point charge at a distance r at point P. For this we divide the total distance from point P to infinity in equal small element of length (dx) i.e PA = AB = BC = CD -------∞
A test charge +qo be placed at a distance x from this point charge.
By coulomb law the repulsion force exerted by +q charged on test charge +qo which is placed at a distance x is---
F = 1/4πεo × q·qo/x² N
small work done against this repulsion Force for bringing a small distance dx will be dw
dw = F·dx·cosθ
putting the values
dw = (1/4πεo × q·qo/x ) dx cos(180)
dw = (1/4πεo × q·qo/x) dx (-1)
dw = - (1/4πεo × q·qo/x) dx
taking integration of both side in the limit ∞------r
∫dw = ∫- (1/4πεo × q·qo/x) dx
“Electric potential”:
Electric potential is the physical quantity by which we can determine the direction of flow of charge. It is represented by the “V”.
*Measurment: Electric potential at a point in an electric field is equal to the work done in bringing a unit positive test charge from infinity to that point.
Electric potential, V = W/q•
Units :
1) SI unit = joule/coulomb or volt
2) CGS unit = erg/stat-coulomn or stat-volt
Dimensional
Formula = [ML^2T^-3A^-1]
*Relationship between volt and stat-volt ---
(Electric
potential )= work done/charge
1 volt = 1 joule/ 1coulomb
= 10^7 erg/ 3×10^9 stat-coulomb
1 volt = 1/300 stat-volt
2nd topic
*“Potential difference”: It is the work done in bringing a unit positive test charge from one point to another point. It is represented by V or ΔV.
It is scalar quantity.
According to definition ----
V2 - V1 = W/ q•
ΔV = (F. S. CosΦ) / q•
[ here Φ = 180 and F = q. E ]
ΔV = [(q• E)ΔX. CosΦ]/ q•
ΔV = E. ΔX. (-1)
ΔV = - E. Δx
E = - ΔV/ ΔX
[ E = - dv/ dx ]
Electric field intensity = negative potential gradient
3rd topic:
*“Electric potential due to point charge”:
In above figure a point charge +q be placed at point 0. We are to find the electric potential due to this point charge at a distance r at point P. For this we divide the total distance from point P to infinity in equal small element of length (dx) i.e PA = AB = BC = CD -------∞
A test charge +qo be placed at a distance x from this point charge.
By coulomb law the repulsion force exerted by +q charged on test charge +qo which is placed at a distance x is---
F = 1/4πεo × q·qo/x² N
small work done against this repulsion Force for bringing a small distance dx will be dw
dw = F·dx·cosθ
putting the values
dw = (1/4πεo × q·qo/x ) dx cos(180)
dw = (1/4πεo × q·qo/x) dx (-1)
dw = - (1/4πεo × q·qo/x) dx
taking integration of both side in the limit ∞------r
∫dw = ∫- (1/4πεo × q·qo/x) dx
w = qqo/4πεor
w/qo = 1/4πεo × q/r
w/qo = 1/4πεo × q/r
[According to the definition
electric potential V =W/qo ]
∴ [V = 1/4πεo × q/r ] volt
In CGS system 1/4πεo = 1
∴ [ V = q/r ] stat-volt
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