Electric Potential And Capacitance

Electric potential is the potential energy per unit charge in an electric field, influenced by point charges, dipoles, and charged spheres. Capacitance is a conductor's ability to store charge, depending on its geometry and surrounding dielectrics. Capacitors store energy and can be combined in series or parallel to adjust total capacitance.e

1.0Definition Electric Potential

It is the work done by an external force to move a unit positive charge from a reference point to a given point without changing its kinetic energy.

$V_p=\frac{(W_{\infty \to p}{)}_{ext}}{q}$        ($\Delta K=0$)

• It is scalar quantity
• Potential can be positive, negative and even zero
• SI unit is joule/coulomb

2.0Electric Potential Due To a Point Charge

Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field, without changing its kinetic energy.

Potential due to a positive point charge

$V_P=\frac{kQ}{r}$

Potential due to a negative point charge

$V_P=-\frac{kQ}{r}$

Note: Where reference potential is 0 at infinity.

3.0Electric Potential Due to System of Charges

$V=V_1+V_2+V_3+........V_n$

$V_1=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r_{1P}}$ , $V_2=\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{r_{2P}}$, $V_3=\frac{1}{4\pi {\epsilon }_0}\frac{q_3}{r_{3P}}$, $V_n=\frac{1}{4\pi {\epsilon }_0}\frac{q_n}{r_{nP}}$

4.0Potential Due to an Electric Dipole

1. At axial point

Electric potential due to +q charge,$V_1=\frac{kq}{r-l}$

Electric potential due to –q charge, $V_2=-\frac{kq}{r+l}$

Net electric potential $V=V_1+V_2=\frac{kq\times 2l}{r^2-l^2}$

If $r>>l,V=\frac{kp}{r^2}$

2. At equatorial point

Electric potential at P due to +q charge, $V_1=\frac{kq}{x}$

Electric potential at P due to -q charge, $V_1=-\frac{kq}{x}$

Net electric potential $V=V_1+V_2=0$

3. Potential at general point (r, $\theta$) 

So, net potential at point A, $V=\frac{k(pcos\theta }{r^2}$

$V=\frac{k(\vec{p}.\vec{r})}{r^3}Here\vec{r}=\overrightarrow{OA}$

5.0Potential due to a Uniformly Charged Ring or Circular Arc

1. At the centre

$V=\frac{kQ}{R}$

2. On its axis 

$V=\frac{Q}{4\pi {\epsilon }_0(R^2{+}^2{)}^{\frac{1}{2}}}=\frac{kQ}{\sqrt{R^2+x^2}}=\frac{kQ}{r}$

3. Potential due to a Uniformly Charged Circular Arc

$V=\frac{kQ}{R}=\frac{k(\theta R\lambda )}{R}=k\theta \lambda$

6.0Potential Due to a Uniformly Charged Sphere

1. Electric Potential due to a Charged Conducting Sphere 

a. For outside points (r > R)

           $V=\frac{kQ}{r}=\frac{\sigma R^2}{{\epsilon }_{0}r}$

b. For points on the surface (r = R)

           $V=\frac{\sigma R}{{\epsilon }_0}=\frac{kQ}{R}$

c. For Inside points (r < R)

            $V=\frac{kQ}{r}$

Variation of potential with r

2. Electric Potential due to Uniformly Charged Non-conducting Sphere

a. Outside (r > R)

   $V=\frac{kQ}{r}$

b. On the surface (r = R)

   $V=\frac{kQ}{r}$

c. Inside (r < R)

$V_{inside}=\frac{kQ}{2R^3}[3R^2-r^2]$

Variation of potential with r

7.0Relationship Between Electric Field and Electric Potential

$F=-\frac{dU}{dr}$

Dividing both sides by q

$\frac{F}{q}=-\frac{d}{dr}(\frac{U}{q})\Rightarrow E=-\frac{dV}{dr}(F=qEandU=qV)$

$\vec{E}=-\overrightarrow{▽}V$

$E_x=-\frac{\partial V}{\partial x}$,    $E_y=-\frac{\partial V}{\partial y}$, $E_z=-\frac{\partial V}{\partial z}$

8.0Equipotential Surfaces

The locus of all such points which are at same potential or the surface at which potential of all the points are equal is known as equipotential surface (EPS). It can be of any shape.

Properties of equipotential surface

•  Electric field intensity is always perpendicular to EPS.
•  Work done in moving a charge from one point of EPS to another point of the same EPS   is always zero.
•  Two EPS can never intersect each other.

1. EPS due to point charge - EPS of point charge is spherical in shape.

2. EPS due to line charge – EPS of line charge is cylindrical in shape.

3. EPS due to large plane charged sheet – EPS of a large plane sheet is plane in shape.

4. EPS due to unlike charges -

5. EPS due to like charges -

9.0Electric Potential Energy of a System of Charges

Electrostatic potential energy is the work done to bring a charge from infinity to its position without changing its kinetic energy.

Electrostatic Potential Energy (EPE) of two point charge system 

$\Delta U=(W_{\infty \to r}{)}_{ext}=-{\int }_{\infty }^r{\vec{F}}_{ext}.\overrightarrow{dr}$

$U_r-U_{infty}=-{\int }_{\infty }^r{\vec{F}}_{ext}.\overrightarrow{dr}=\frac{kQq}{r}$

$(\therefore U_{\infty }=0)$

$U-0=\frac{kQq}{r}$

$U=\frac{kQq}{r}$

10.0Electrostatic Potential energy of a three system of charge

$U_S=U_{12}+U_{23}+U_{13}$

$U_S=\frac{k{q}_1{q}_2}{r_{12}}+\frac{k{q}_2{q}_3}{r_{23}}+\frac{k{q}_1{q}_3}{r_{13}}$

Note: For n point charges system 

No. of pairs= $\frac{n(n-1)}{2}$

11.0Electrostatics of Conductors

1. Electric Field Inside a Conductor: The electric field inside a conductor in electrostatic equilibrium is zero.
2. Charge Distribution: Any excess charge on a conductor resides entirely on its surface.
3. Equipotential Surface: The entire conductor (including its surface) is at the same potential in electrostatic equilibrium.
4. Electric Field Just Outside the Surface: The field is perpendicular to the surface.
5. No Electric Field Parallel to the Surface: If there were, charges would move, contradicting equilibrium.
6. Cavity Inside a Conductor (No Charge):The electric field inside a cavity with no charge is zero, even if the conductor is charged.
7. Cavity with Internal Charge: Induced charges appear on the inner surface of the cavity to cancel the field inside the conductor.
8. Grounding: Connecting a conductor to the Earth allows excess charge to flow, maintaining zero potential.

12.0Properties of Conductors in Electrostatic Equilibrium

• Conductors are materials which contain a large number of free electrons which can move freely inside the conductor.
• Electrostatic field lines never exist inside a conductor.
• Electric field lines terminate & originate always perpendicular to the surface of conductor because the surface of conductor is an equipotential surface.
• Charge always resides on the outer surface of a conductor.
• If there is a cavity inside a charged conductor with the cavity devoid of any charge then charge will always reside only on the outer surface of the conductor.
• Electric field intensity near a conducting surface is given by,$\vec{E}=\frac{\sigma }{{\epsilon }_0}\hat{n}$
• When a conductor is grounded (earthed), its electric potential becomes zero. However, this doesn't always mean its charge is zero—only if the body is isolated can grounding ensure the charge is also zero.

13.0Electrostatic Shielding

• It is the method of protecting a certain region from the effect of an electric field.
•  A cavity surrounded by conducting walls is a field free region as long as there are no charges inside the cavity.
•  Whatever be the charge and field configuration the field inside the cavity is always zero (Provided no charge present inside the cavity) This is known as electrostatic shielding.
•  Electrostatic shielding can be achieved by enclosing sensitive instruments in a hollow conductor.
• Figure gives a summary of the important electrostatic properties of a conductor.

14.0Capacitance of a Conductor

It demonstrates a conductor's ability to store electrical energy through an electric field. If charge(Q) is given to an isolated conducting body and it's potential increases by V, then

$\Rightarrow$Q ∝ V              , Q =CV

$\Rightarrow =\frac{Q}{V}$ (C= Capacitance of the capacitor)

• Electrical capacitance is a Scalar quantity.
• Capacitance of conductor depends upon shape, size, presence of medium and nearness of other conductor.

Graph Between Q and V

• S I Unit- Farad
• CGS Unit-Stat Farad
• Dimensional Formula-[M^-1L^-2T⁴A²]
• Combination of Capacitors

15.0Combination of Capacitors

1. Series Combination

• Capacitors are connected end-to-end so that the same current flows through each Capacitor. The total Capacitance in series is less than any individual capacitor's Capacitance. The Charge on each Capacitor connected in series is the same.
•  When two capacitors   are connected in series, then effective capacitance is given by                     

                        $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$  

                        $\frac{1}{C}=\frac{C_1+C_2}{C_1{C}_2}$   

                        $C=\frac{C_1{C}_2}{C_1+C_2}$

• The effective capacitance of capacitors in series connection is lower than the capacitance of each capacitor individually.
• The Charge for each capacitor in the series is the same.

2. Parallel Combination of Capacitor

• Capacitors are connected across each other's terminals and share the same voltage. When a potential difference V is applied across the terminals all capacitors have equal potential difference. The equivalent Capacitance of parallel combination is more significant than any of the capacitances in the combination.
• Effective Capacitance is given by, C=C₁+C₂+C₃
• Effective Capacitance of parallel combination is greater than any of the capacitance.
• In Parallel combination, voltage across each Capacitor is the same.

Parallel Plate Capacitor(PPC)

• It comprises two large, flat, parallel conducting plates separated by a small distance.
• $C=\frac{{\epsilon }_{0}A}{d}$
• Capacitance of parallel plate capacitor with dielectric

$C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}={\epsilon }_r{C}_0(\therefore C_0=\frac{{\epsilon }_{0}A}{d})$

Capacitance of Parallel Plate Capacitor Depends on

• Area $\Rightarrow C\propto A$
• Distance between the plates $\Rightarrow C\propto \frac{1}{d}$
• Medium between the Plates $\Rightarrow C\propto {\epsilon }_r$

Capacitance of Isolated Spherical Conductor

• If the medium around the conductor is vacuum or air than capacitance is given by, $C=4\pi {ϵ}_{0}R$ (∴R = Radius of Spherical Conductor(Solid or Hollow)
• If the medium around the conductor is a dielectric of constant K from surface of sphere to infinity then $C_{medium}=4\pi {\epsilon }_{0}KR$
• $\frac{C_{medium}}{C_{air/vacuum}}=K$=Dielectric Constant

Capacitance of Spherical Capacitor

Case 1: Outer Sphere is Earthed

• When a charge Q is given to the inner sphere it is uniformly distributed on its surface A charge –Q is induced on the inner surface of the outer sphere. The charge +Q induced on the outer surface of the outer sphere flows to earth as it is grounded.
• E=0 For r<a and E=0 for r>b ,this arrangement is known as spherical capacitor
• Capacitance of spherical capacitor is given by $C=\frac{4\pi {\epsilon }_{0}ab}{b-a}$
• If dielectric medium is filled then $C=\frac{4\pi {\epsilon }_0{\epsilon }_{r}ab}{b-a}$

Case 2: Inner Sphere is earthed 

Capacitance is given by, $C=\frac{4\pi {\epsilon }_0{b}^2}{b-a}$

Cylindrical Capacitor

There are two coaxial conducting cylindrical surfaces where l  >> a and l >> b, where a and b are the radius of cylinders. When a charge Q is given to the inner cylinder it is uniformly distributed on its surface. A charge –Q is induced on the inner surface of the outer cylinder. The charge +Q induced on outer surface of outer cylinder flows to earth as it is grounded. Hence Capacitance is given by                  $C=\frac{2\pi {\epsilon }_{0}l}{ln(\frac{b}{a})}$

16.0Dielectrics and Their Effect on Capacitance

Capacitors with Dielectric

1. Electric Field in the absence of Dielectric, $E=\frac{\sigma }{{\epsilon }_0}$
2. Electric Field in the absence of Dielectric,$E_0=\frac{Q}{{\epsilon }_{0}A}$
3. Electric Field in absence of Dielectric, $E=\frac{V}{d}$
4. Capacitance in absence of Dielectric, $C_0=\frac{Q}{V_0}$
5. Capacitance in presence of Dielectric, $C=\frac{Q-Q_b}{V}$
6. Dielectric Constant or Relative Permittivity (K or ${\epsilon }_r$)

$K=\frac{ϵ}{{ϵ}_0}=\frac{E_0}{E}=\frac{V_0}{V}=\frac{C}{C_0}=\frac{Q}{Q-Q_b}=\frac{\sigma }{\sigma -b}$

In presence of dielectric, capacitance is increased by a factor K

$C=\frac{{ϵ}_{0}AK}{d}=K{C}_0$

Capacitance of Parallel plate Capacitor when dielectric is partially filled

Capacitance is given by, $C=\frac{{ϵ}_{0}A}{(d-t)+\frac{t}{{ϵ}_r}}$

17.0Effects of Dielectrics in Capacitor

1. Distance Division

• Distance is Divided and area remains same
• Capacitors are in Series
• Individual Capacitance are $C_1=\frac{{ϵ}_0{ϵ}_{r1}A}{d_1}$  and   $C_2=\frac{{ϵ}_0{ϵ}_{r2}A}{d_2}$

• Effective Capacitance is given by, $C={ϵ}_{0}A[\frac{{ϵ}_{r1}{ϵ}_{r2}}{d_1{ϵ}_{r2}+d_2{ϵ}_{r1}}]$
• Special Case: If $d_1=d_2=\frac{d}{2}\Rightarrow C=\frac{{ϵ}_{0}A}{d}[\frac{2{ϵ}_{r1}{ϵ}_{r2}}{{ϵ}_{r1}+{ϵ}_{r2}}]$

2. Area Division

• Area is divided and distance remains same
• Capacitors are in Parallel
• Individual capacitors are

$C_1=\frac{{ϵ}_0{ϵ}_{r1}{A}_1}{d}$           $C_2=\frac{{ϵ}_0{ϵ}_{r2}{A}_2}{d}$

• Effective Capacitance is given by $C=\frac{{ϵ}_0{ϵ}_{r1}{A}_1}{d}+\frac{{ϵ}_0{ϵ}_{r2}{A}_2}{d}$
• Special Case: If $A_1=A_2=\frac{A}{2}\Rightarrow C=\frac{{ϵ}_{0}A}{d}(\frac{{ϵ}_{r1}+{ϵ}_{r2}}{2})$

18.0Energy Stored in a Capacitor

A capacitor is a device designed to store electrical energy. The process of charging a capacitor entails transferring electric charges from one plate to another. The work done during this charging process is stored as electrical potential energy within the capacitor.

$V=\frac{W}{q}=\frac{dW}{dq}$

$dW=Vdq=\frac{q}{C}dq$

$W=\int dW={\int }_0^Q\frac{q}{C}dq=\frac{1}{C}\frac{Q^2}{2}=\frac{Q^2}{2C}$

$U=\frac{Q^2}{2C}=\frac{1}{2}C{V}^2=\frac{1}{2}QV$

19.0Sample Questions on Electric Potential and Capacitance

Q-1. A conducting sphere of radius R has electric field E at distance 3R from the centre. Find potential difference between the centre of the sphere & the point where the electric field is E.

Solution:

$E=\frac{kQ}{(3R{)}^2}=\frac{kQ}{9R^2}\Rightarrow kQ=9E{R}^2$

$V_{Centre}=\frac{kQ}{R}=\frac{9E{R}^2}{R}=9ER$

$V_P=\frac{kQ}{r}=\frac{9E{R}^2}{3R}=3ER$

$V_{Centre}-V_P=9ER-3ER=6ER$

Q-2. Find potential at points A,B and C due to the given arrangement of shells.

Solution:

$V_A=\frac{k{Q}_1}{a}+\frac{k{Q}_2}{b}$

$V_B=\frac{k{Q}_1}{a}+\frac{k{Q}_2}{b}$

$V_C=\frac{k{Q}_1}{b}+\frac{k{Q}_2}{b}$

Q-3. A charge in moving via three paths, 1, 2, and 3. as shown in figure. Find work done in each case.

Solution: Potential at points A, B, C and D is $V=\frac{kq}{R}$. All the points are equipotential points, so work done in each case is zero.

Q-4. Calculate potential due to a charge $5\times 10^{-7}$ C at a point P located 9 cm away. Also find work done in bringing an another charge of $3\times 10^{-8}$ C from infinity to the point.

Solution:

$V_P=\frac{kq}{r}=\frac{9\times 10^9\times 5\times 10^{-7}}{9\times 10^{-2}}=5\times 10^{4}V$

$W=q(\Delta V)=q(V_P-V_{infty})=3\times 10^{-8}(5\times 10^4-0)=15\times 10^{-4}=1.5mJ$

Q-5. Find p.d. between points A and B as shown in figure.

Solution:

$V_A-V_B=2k\lambda ln(\frac{r_B}{r_A})$

$V_A-V_B=2k\lambda ln(\frac{2a}{}a)$

$V_A-V_B=2k\lambda ln(2)$