keywords- ||class 12 physics chapter 1 notes || class 12 physics chapter 2 electrostatic potential and capacitance notes ||
- Electrostatic Potential Energy, Electrostatic potential
- Electrostatic Potential due to a Dipole, Equipotential Surfaces, Electrostatic Potential Energy of a system of Charges
- Capacitor and Capacitance
- Grouping of Capacitors
- Energy stored in a capacitor, Common Potential, Loss of Energy in Sharing Charges
Electric Potential Difference
A potential difference between two points in an electric field may be defined as the amount of work done in moving a unit of positive charge from one point to the other against the electrostatic forces.
Figure
If V_{P} and V_{R} are the electric potential at points P and R respectively, then △V=V_{P}−V_{R} or \triangle\;V=\frac{W_{PR}}{q} and V_{P}−V_{R}=\frac{W_{PR}}{q}
- SI unit of potential difference \frac{J}{C}\;or\;JC^{−1}or\;volt
Note: 1. Test charge is so small that it does not disturb the distribution of the source charge. 2. We just apply so much external force on the day chart that is just becomes the repulsive electric force on it and hence does not produce any acceleration in it. (i.e. Potential difference is path independent)
1 Volt Potential Difference
We know V_{A}−V_{B}=\frac{W}{q} If W = 1 joule and q = 1 coulomb then V_{A}−V_{B}=1\;volt So we can say that, if 1-joule work is done in taking a test charge of 1 coulomb from one point to another in an electric field, then the potential difference between these two points will be 1 volt.
Electric Potential
The electric potential at any point in an electric field is the amount of work done in moving a unit positive charge from infinity to that point against the electrostatic forces.
If the point R lies at infinity, then V_{R}=0 so that, V=V_{P}=\frac{W}{q}\;\;\;\;or\;\;\;V=\frac{W}{q}
- Electric potential decreases in the direction of the electric field.
- Electric potential is a scalar quantity.
- \;Dimension\;\;\colon\;\lbrack\;ML^{2}T^{−3}A^{−1}\;\rbrack\;
1.Electric Potential Due to a Point Charge
Consider a +ve point charge q placed at the origin O. We have to calculate the potential at point P.
Let a test charge is placed at point A.
According to Coulomb's Law, Force acting on charge q_{o} is F=\frac{1}{4πε_{o}K}\frac{qq_{o}}{x^{2}}....(1) Work done in moving the test charge q_{o} From A to B through small displacement d\overrightarrow{x} against the electric force is - dW=\overrightarrow{F}\;·d\overrightarrow{x}\; dW=Fdx\;cos180^{\;∘\;} dW=−Fdx The total work done in moving the charge q_{o} from infinity to that point P will be W=\int_{\infty\;}^{r}dW W=\int_{\infty\;}^{r}−Fdx W=−\int_{\infty\;}^{r}Fdx W=−\int_{\infty\;}^{r}\frac{1}{4πε_{ο}}\frac{qq_{ο}}{x^{2}}dx ......By eqn (1) W=−\frac{qq_{ο}}{4πε_{ο}}\int_{\infty\;}^{r}x^{−2}dx W=−\frac{qq_{ο}}{4πε_{ο}}\;\lbrack\;\frac{x^{−2+1}}{−2+1}\;\rbrack\;_{\infty\;}^{r} W=−\frac{qq_{ο}}{4πε_{ο}}\;\lbrack\;−\frac{1}{x}\;\rbrack^{r}_{\infty\;} W=\frac{qq_{ο}}{4πε_{ο}}\;\lbrack\;\frac{1}{r}−\frac{1}{\;\infty\;}\;\rbrack\; W=\frac{1}{4πε_{ο}}\frac{qq_{ο}}{r} .............(2) Hence electric potential V=\frac{W}{q_{ο}} V=\frac{qq_{ο}}{4πε_{ο}r}×\frac{1}{q_{ο}} ......By eqn (2) so, [ V=\frac{1}{4πε_{ο}}\frac{q}{r} ]
2. Electric Potential due to a Dipole:
(A) Electric Potential due to Dipole at a Point on Axial Line
Potential due to +q and – q charges at P is
V_{+}=\frac{1}{4πε_{ο}}\frac{q}{\;\lparen\;r−a\;\rparen}
V_{−}=−\frac{1}{4πε_{ο}}\frac{q}{\lparen\;r+a\;\rparen}
The total potential at point P is
V=V_{+}+V_{−} V=\frac{1}{4πε_{ο}}\frac{q}{\;\lparen\;r−a\;\rparen}−\frac{1}{4πε_{ο}}\frac{q} {\;\lparen\;r+a\;\rparen\;\;} V=\frac{q}{4πε_{o}}\lbrack\frac{1}{\;\lparen\;r−a\;\rparen}−\frac{1}{\;\lparen\;r+a\;\rparen}\;\rbrack\; V=\frac{q}{4πε_{ο}}\;\lbrack\;\frac{\;\lparen\;r+a\;\rparen\;−\;\lparen\;r−a\;\rparen\;}{\;\lparen\;r+a\;\rparen\;\;\lparen\;r−a\;\rparen\;}\;\rbrack\; V=\frac{q}{4πε_{ο}}\;\lbrack\;\frac{r+a−r+a}{r^{2}−a^{2}}\;\rbrack V=\frac{q}{4πε_{o}}\;\lbrack\;\frac{2a}{r^{2}−a^{2}}\;\rbrack V=\frac{1}{4πε_{ο}}\frac{2qa}{r^{2}−a^{2}} hence p=q.2a V=\frac{1}{4πε_{o}}\frac{p}{r^{2}−a^{2}}\; for short dipole r>>>a, so we can neglect 'a' for very short dipole so, [ V_{axial}=\frac{1}{4πε_{ο}}\frac{p}{r^{2}} ]
(B) Potential Due to Dipole at a Point on Equitorial Line
Potential due to +q and – q charges at P is
V_{+}=\frac{1}{4πε_{ο}}\frac{q}{\;\lparen\;r^{2}+a^{2}\;\rparen\;^{\frac{1}{2}}} V_{-}=\frac{1}{4πε_{ο}}\frac{-q}{\;\lparen\;r^{2}+a^{2}\;\rparen\;^{\frac{1}{2}}} So, the total potential at point P is V_{eq}=V_{+}\;+V_{−} [ V_{eq}=0 ]
(C) Potential Due to Dipole at any Point
Potential due to +q and -q at point P are respectively
V_{+}=\frac{1}{4πε_{ο}}\frac{q}{r_{1}} V_{−}=\frac{1}{4πε_{ο}}\frac{−q}{r_{2}} So, total potential at P V=V_{+}+V_{−} V=\frac{1}{4πε_{ο}}\frac{q}{r_{1}}−\frac{1}{4πε_{ο}}\frac{q}{r_{2}} V=\frac{q}{4πε_{ο}}\;[\;\frac{1}{r_{1}}−\frac{1}{r_{2}}\;\rbrack\; V=\frac{q}{4πε_{ο}}\;\lbrack\;\frac{r_{2}−r_{1}}{r_{1}r_{2}}\;\rbrack\; If the point P lies far away from the dipole, then r_{1}−r_{2}≃2acosθ\;and\;r_{1}r_{2}≃r^{2} So, V=\frac{q}{4πε_{ο}}×\frac{2a\;Cosθ}{r^{2}} V=\frac{1}{4πε_{ο}}\frac{2qa\;Cosθ}{r^{2}} [V=\frac{1}{4πε_{o}}\frac{pcosθ}{r^{2}}]\;\;∵\;p=2qa
(D) Potential Due to a System of Charges
Let there be n number of charges q_{1\;,}q_{2}\;,q_{3}…\;…\;q_{n} at distances r_{1\;,}r_{2}\;,r_{3}…\;…\;r_{n} from point P.
Potential due to q_{1}\;and\;q_{2} charges at P is given by V_{1}=\frac{1}{4πε_{ο}}\frac{q_1}{r_{1p}} V_{2}=\frac{1}{4πε_{ο}}\frac{q_2}{r_{2p}} . . . Similarly V_{n}=\frac{1}{4πε_{ο}}\frac{q_n}{r_{np}} The total potential at point P due individual charges is given by V=V_{1}+V_{2}+V_{3}+\;…\;+V_{n} V=\frac{1}{4πε_{ο}}\frac{q_1}{r_{1p}} + \frac{1}{4πε_{ο}}\frac{q_2}{r_{2p}} +............. \frac{1}{4πε_{ο}}\frac{q_n}{r_{np}} So, V=\frac{1}{4πε_{ο}}\sum_{i=1}^{n}\frac{q_{i}}{r_{iP}}
Equipotential Surface:
Any surface that has the same potential at every point on it is called an “Equipotential Surface”. An equipotential surface is an imaginary surface.
Properties of Equipotential Surface
1.No work is done in moving a test charge over an equipotential surface.
- ∵W=q_{ο}\;\lparen\;V_{A}−V_{B}\;\rparen on equatorial surface V_{A}−V_{b}=0 so W=0
2.The Electric Field is always perpendicular (Normal) to the equipotential surface at every point.
- The field were not normal to the equatorial surface, it would have a non zero component along the surface. So to move a test charge against this component, a work done have to be done
3.Equipotential Surfaces are closer together in the region of strong fields and farther apart in the region of weak fields.
- Because E=−\frac{dV}{dr} or dr=−\frac{dV}{E} on equitorial surface dV=Constant then r\;∝\;\frac{1}{r}
4.No two equipotential Surfaces can intersect each other. because at the point of intersection there will be two values of electric potential at the point of intersection, which is impossible.
Relation Between Field and Potential:
Let us consider two equipotential surfaces A & B. Let P be a point on the surface B and δl is the perpendicular distance of the Surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the Surface B to surface A against the electric field.
The work done in this process is - dW=\overrightarrow{F}_{eF}\;•\;δ\overrightarrow{l} here \overrightarrow{F}_{eF}=−\overrightarrow{F}_{ext} (∵external force is against electrostatic force) so, dW=−\overrightarrow{F}_{ext}\;•\;δ\overrightarrow{l} dW=−q_{ο}\overrightarrow{E}\;•\;δ\overrightarrow{l} \frac{dW}{q_{ο}}=−\overrightarrow{E}\;•\;δ\overrightarrow{l} dV=−\overrightarrow{E}\;•\;δ\overrightarrow{l} ----> For uniform electric field and -ve shiwing that potential always decreased in the direction of electric field. For Non-uniform electric field V=−\int_{A}^{B}\overrightarrow{E}\;•\;δ\overrightarrow{l} Or E=−\frac{dV}{δl}
Potential Energy of a System of Charges:
It is equal to the total amount of work done in assembling all charges at a given position from infinity.
In bringing the first charge {q}_{1} to\;position\;{P}_{1}(\overrightarrow{r}_{1}) , no work is done, because all other charges are still at infinity. & there is no field.
i.e. W_{1}=0
When we bring charges q_{2} from infinity to {P}_{2}(\overrightarrow{r}_{2}) from {q}_{1} work is done
W_{2}= Potential\;due\;to\;charge\;q_{1}\;×q_{2} W_{2}=\frac{1}{4πε_{o}}\frac{q_{1}}{r_{12}}×q_{2}\;\;\;\;\;∵V_{1}\;=\frac{1}{4πε_{ο}}\frac{q_{1}}{r_{12}} In bringing q_{3} from infinity to {P}_{3}(\overrightarrow{r}_{3}) atwork has to be done against electrostatics forces of both q_{1} & q_{2} W_{3}=[ potential\;due\;to\;q_{1}\;and\;q_{2}]×\;charge\;q_{3} W_{3}=\lbrack\frac{1}{4πε_{ο}}\frac{q_{1}}{r_{13}}+\frac{1}{4πε_{ο}}\frac{q_{2}}{r_{23}}\rbrack×q_{3}\; W_{3}=\frac{1}{4πε_{ο}}\;\lbrack\;\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\;\rbrack\; Potential energy of the three charges = Total work done to bring all th ethree charges U=W_{1}+W_{2}+W_{3} U=0+\frac{1}{4πε_{o}}\frac{q_{1}q_{2}}{r_{12}}+\frac{1}{4πε_{ο}}\;\lbrack\;\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{24}}\;\rbrack\; U=\frac{1}{4πε_{o}}\;\lbrack\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{24}}\rbrack\; -----> This formula is valid only, when there is no external electric field.
Note 1. Potential Energy of a system of one charges when there is no external electric field.
U = 0
Note 2. Potential Energy of a system of two charges when there is no external electric field.
U_{2}=\frac{1}{4πε_{o}}\frac{q_{1}q_{1}}{r_{12}}
Potential Energy of Charges in an External Electric Field:
1. Potential Energy of a Single Charges
