Electrostatic Potential and Capacitance Notes

Introduction to Electrostatic Potential

• Potential energy was introduced (Chapters 5 and 7, Class XI).

• Work done by an external force against a conservative force (spring or gravity) is stored as potential energy.

• When the external force is removed, potential energy converts to kinetic energy, conserving the total energy.

• Examples of conservative forces: spring force, gravitational force, and Coulomb force.

• Coulomb force is conservative due to its inverse-square dependence on distance, similar to gravitational force.

Electrostatic Potential Energy

• Analogous to gravitational potential energy, electrostatic potential energy can be defined for a charge in an electrostatic field.

• Consider an electrostatic field $E$ due to a charge configuration.

• For simplicity, consider $E$ due to a charge $Q$ at the origin.

• Imagine bringing a test charge $q$ from point $R$ to point $P$ against the repulsive force from $Q$ (assuming both are positive).

• Assume $q$ is small enough not to disturb the original configuration (or $Q$ is held fixed).

• Apply an external force F{ext}$to counter the electric force$FE$($F{ext} = -FE$).</p></li><li><p>The charge$q$is brought from$R$to$P$with infinitesimally slow constant speed (no net force or acceleration).</p></li><li><p>Work done by$F{ext}$is the negative of work done by$FE$and is stored as potential energy of$q$.</p></li><li><p>If$F{ext}$is removed at$P$,$FE$will move$q$away from$Q$, converting potential energy to kinetic energy.</p></li></ul><h4 id="a3e04dc0-a267-4200-a805-5141ca680f6d" data-toc-id="a3e04dc0-a267-4200-a805-5141ca680f6d" collapsed="false" seolevelmigrated="true">Work Done and Potential Energy Difference</h4><ul><li><p>Work done by external forces in moving a charge$q$from$R$to$P$is described by the equation:$W{RP} = -\int{R}^{P} F_E \cdot dl$</p></li><li><p>$W_{RP}$is against the electrostatic repulsive force and is stored as potential energy.</p></li><li><p>A charge$q$in an electric field possesses electrostatic potential energy.</p></li><li><p>The potential energy difference between points$R$and$P$is:$\Delta U = UP - UR = W_{RP}

• Displacement is opposite to the electric force, so work done by the electric field is negative ($-W_{RP}).

• Electric potential energy difference is the work done by an external force in moving charge $q$ from one point to another without acceleration.

Comments on Potential Energy

• (i) The right side of the equation depends only on the initial and final positions of the charge; the electrostatic field's work depends only on the initial and final points, independent of the path.

• This path-independence is a fundamental characteristic of conservative forces.

• Potential energy is meaningful only if work is path-independent.

• Path-independence can be proved using Coulomb’s law (proof omitted).

• (ii) Potential energy difference is defined in terms of work, a physically meaningful quantity.

• Potential energy is undetermined to within an additive constant, meaning the actual value is not physically significant, only the difference is.

• Adding a constant $\alpha$ to potential energy at every point does not change the potential energy difference: $(U<em>P + \alpha) - (U</em>R + \alpha) = U<em>P - U</em>R$

• There is freedom in choosing the point where potential energy is zero.

• A convenient choice is to have electrostatic potential energy zero at infinity.

• With this choice, setting point $R$ at infinity, we get: $U<em>P = W</em>{\infty P}$

• Potential energy of charge $q$ at a point is the work done by the external force in bringing $q$ from infinity to that point.

Electrostatic Potential

• Electrostatic potential is characteristic of the electric field.

• Work done per unit test charge is independent of $q$.

• From Eq. (2.1): $V<em>P - V</em>R = \frac{U<em>P - U</em>R}{q}$

• $V<em>P$ and $V</em>R$ are electrostatic potentials at $P$ and $R$, respectively.

• Only potential difference is physically significant.

• Choosing potential to be zero at infinity:

• Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V) at that point.

• Mathematically, $V = \frac{W_{\infty P}}{q}$

• To obtain work done per unit test charge, use an infinitesimal test charge $\delta q$, obtain work done $\delta W$ in bringing it from infinity, and determine the ratio $\frac{\delta W}{\delta q}$

• The external force at every point must be equal and opposite to the electrostatic force.

Potential Due to a Point Charge

• Consider a point charge $Q$ at the origin (take $Q$ > 0).

• Determine the potential at point $P$ with position vector $r$ from the origin.

• Calculate the work done in bringing a unit positive test charge from infinity to point $P$.

• For $Q$ > 0, work done against the repulsive force on the test charge is positive.

• Work done is independent of the path; choose a convenient radial path from infinity to $P$.

• At an intermediate point $P'$ on the path, the electrostatic force on a unit positive charge is:
  $\frac{1}{4 \pi \epsilon_0} \frac{Q}{r'^2} \hat{r}'$

• Work done against this force from $r'$ to $r' + \Delta r'$ is:
  $\Delta W = - \frac{Q}{4 \pi \epsilon_0} \frac{\Delta r'}{r'^2}$

• The negative sign appears because for \Delta r' < 0, $\Delta W$ is positive.

• Total work done ($W$) by the external force is obtained by integrating:
  $W = - \int<em>{\infty}^{r} \frac{Q}{4 \pi \epsilon</em>0} \frac{dr'}{r'^2} = \frac{Q}{4 \pi \epsilon_0 r}$

• By definition, this is the potential at $P$ due to the charge $Q$:
  $V(r) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$

Sign of Charge and Potential

• Equation is true for any sign of $Q$, though we considered $Q$ > 0 in its derivation.

• For $Q$ < 0, $V$ < 0, i.e., work done (by the external force) per unit positive test charge is negative.

• The work done by the electrostatic force in bringing the unit positive charge from infinity to $P$ is positive.

• Since for $Q$ < 0, the force on a unit positive test charge is attractive, the electrostatic force and the displacement (from infinity to $P$) are in the same direction.

• Equation is consistent with the choice that potential at infinity be zero.

Variation of Potential and Field with Distance

• Electrostatic potential $(V \propto \frac{1}{r})$ varies inversely with $r$.

• Electrostatic field $(E \propto \frac{1}{r^2})$ varies inversely with $r^2$.

Example 2.1

• (a) Calculate the potential at a point $P$ due to a charge of $4 \times 10^{-7} C$ located 9 cm away.

• (b) Obtain the work done in bringing a charge of $2 \times 10^{-9} C$ from infinity to point $P$.

• Does the answer depend on the path?

Solution:

• (a) $V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r} = (9 \times 10^9 Nm^2C^{-2}) \frac{4 \times 10^{-7} C}{0.09 m} = 4 \times 10^4 V$

• (b) $W = qV = (2 \times 10^{-9} C)(4 \times 10^4 V) = 8 \times 10^{-5} J$
  No, work done is path independent.

The Electric Dipole

• An electric dipole consists of two charges $q$ and $-q$ separated by a small distance $2a$.

• Its total charge is zero.

• A dipole is characterized by a dipole moment vector $p$ whose magnitude is $q \times 2a$ and which points from $-q$ to $q$.

• The electric field depends on the angle between $r$ and $p$.

• The field falls off at a large distance, not as $1/r^2$ (typical of single charge), but as $1/r^3$.

Electric Potential Due to a Dipole

• Electric potential follows the superposition principle.

• The potential due to the dipole is the sum of potentials due to charges $q$ and $-q$:
  $V = \frac{1}{4 \pi \epsilon<em>0} \left( \frac{q}{r</em>1} - \frac{q}{r_2} \right)$

• Where $r<em>1$ and $r</em>2$ are the distances of the point $P$ from $q$ and $-q$ respectively.

• By geometry,
  $r<em>1^2 = r^2 + a^2 - 2ar \cos \theta$ $r</em>2^2 = r^2 + a^2 + 2ar \cos \theta$

• Assuming $r \gg a$,
  $r<em>1 \cong r - a \cos \theta$ $r</em>2 \cong r + a \cos \theta$

• Using the Binomial theorem, and retaining terms up to the first order in $a/r$, we obtain
  $\frac{1}{r<em>1} \cong \frac{1}{r} \left( 1 + \frac{a}{r} \cos \theta \right)$ $\frac{1}{r</em>2} \cong \frac{1}{r} \left( 1 - \frac{a}{r} \cos \theta \right)$

• So, $V = \frac{q}{4 \pi \epsilon_0} \frac{2a \cos \theta}{r^2}$

• Since $p = 2qa$, $V = \frac{1}{4 \pi \epsilon_0} \frac{p \cos \theta}{r^2}$

• Since $p \cos \theta = p.\hat{r}$, electric potential of a dipole is given by
  $V = \frac{1}{4 \pi \epsilon_0} \frac{p.\hat{r}}{r^2}$ , ($r \gg a$).

• Potential on the dipole axis ($\theta = 0, \pi$) is:
  $V = \pm \frac{1}{4 \pi \epsilon_0} \frac{p}{r^2}$

• (Positive sign for $\theta = 0$, negative sign for $\theta = \pi$).

• The potential in the equatorial plane ($\theta = \pi/2$) is zero.

Contrasting Features of Dipole Potential

• (i) The potential depends not just on $r$ but also on the angle between the position vector $r$ and the dipole moment vector $p$.

• (ii) The electric dipole potential falls off, at a large distance, as $1/r^2$, not as $1/r$, characteristic of the potential due to a single charge.

Potential Due to a System of Charges

• Consider a system of charges $q<em>1, q</em>2, …, q<em>n$ with position vectors $r</em>1, r<em>2, …, r</em>n$.

• The potential $V<em>1$ at $P$ due to charge $q</em>1$ is:
  $V<em>1 = \frac{1}{4 \pi \epsilon</em>0} \frac{q<em>1}{r</em>{1P}}$

• $r<em>{1P}$ is the distance between $q</em>1$ and $P$.

• By the superposition principle, the potential $V$ at $P$ is the algebraic sum of the potentials:
  $V = V<em>1 + V</em>2 + … + V_n$

• $V = \frac{1}{4 \pi \epsilon<em>0} \left( \frac{q</em>1}{r<em>{1P}} + \frac{q</em>2}{r<em>{2P}} + … + \frac{q</em>n}{r_{nP}} \right)$

• For a continuous charge distribution characterized by a charge density $\rho(r)$, we have:
  $V = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(r')}{r - r'} dv'$

Uniformly Charged Spherical Shell

• Electric field outside the shell is as if the entire charge is concentrated at the center, so potential outside is:
  $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$, $(r \ge R)$

• q is the total charge and R is its radius.

• Electric field inside the shell is zero.

• Potential is constant inside the shell and equals its surface value:
  $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{R}$

Example 2.2

• Two charges $3 \times 10^{-8} C$ and $-2 \times 10^{-8} C$ are located 15 cm apart.

• Find the point on the line joining the two charges where the electric potential is zero.

Solution:

• Let the origin O be at the positive charge.

• Let P be the required point at x.

• $V = \frac{1}{4 \pi \epsilon_0} \left( \frac{3 \times 10^{-8}}{x} - \frac{2 \times 10^{-8}}{15 - x} \right) = 0$

• $x = 45 cm$

Example 2.3

• Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively.

  • (a) Signs of potential differences $V<em>P – V</em>Q$, $V<em>B – V</em>A$?

  • (b) Sign of potential energy difference of a negative charge between $Q$ and $P; A$ and $B$?

  • (c) Sign of work done by the field moving a positive charge from $Q$ to $P$?

  • (d) Sign of work done by external agency moving a negative charge from $B$ to $A$?

  • (e) Does a negative charge's kinetic energy increase or decrease going from $B$ to $A$?

Solution:

• (a) $V<em>P – V</em>Q$ is positive. $V<em>B – V</em>A$ is Positive.

• (b) Potential energy difference between $Q$ and $P$ is positive and between $A$ and $B$ is positive.

• (c) Work done by the field is negative.

• (d) Work done by external agency is positive.

• (e) Kinetic energy decreases.

Equipotential Surfaces

• An equipotential surface is a surface with a constant value of potential at all points.

• For a single charge $q$, equipotential surfaces are concentric spherical surfaces.

• Electric field lines are radial; the electric field is normal to the equipotential surface.

• In general, the equipotential surface is normal to the electric field.

• If the field were not normal, a non-zero component would exist along the surface, requiring work to move a test charge.

• For a uniform electric field $E$ along the x-axis, equipotential surfaces are planes normal to the x-axis.

Relation Between Field and Potential

• Consider two closely spaced equipotential surfaces $A$ and $B$, with potentials $V$ and $V + \Delta V$.

• $\Delta l$ is the perpendicular distance from $B$ to $A$.

• Moving a unit positive charge from $B$ to $A$ against the field, work done is $|E| \Delta l$.

• $|E| \Delta l = V – (V + \Delta V) = -\Delta V$

• $|E| = - \frac{\Delta V}{\Delta l}$

• Electric field is in the direction of the steepest potential decrease.

• Its magnitude is the change in potential per unit displacement normal to the equipotential surface.

Potential Energy of a System of Charges

• Consider two charges $q<em>1$ and $q</em>2$ with position vectors $r<em>1$ and $r</em>2$.

• Work done to bring $q<em>1$ from infinity to $r</em>1$ is zero.

• The work to bring $q<em>2$ from infinity to $r</em>2$ is $q<em>2$ times the potential at $r</em>2$ due to $q<em>1$. $\frac{1}{4 \pi \epsilon</em>0} \frac{q<em>1 q</em>2}{r_{12}}$

• $r_{12}$ is the distance between points 1 and 2.

• Potential energy of the system is $U = \frac{1}{4 \pi \epsilon<em>0} \frac{q</em>1 q<em>2}{r</em>{12}}$.

• If q1 q2 > 0, potential energy is positive (repulsive force).

• If q1 q2 < 0, potential energy is negative (attractive force).

Generalization for Three Charges

• For three charges $q<em>1, q</em>2$ and $q<em>3$ located at $r</em>1, r<em>2, r</em>3$:$U = \frac{1}{4 \pi \epsilon<em>0} \left( \frac{q</em>1 q<em>2}{r</em>{12}} + \frac{q<em>1 q</em>3}{r<em>{13}} + \frac{q</em>2 q<em>3}{r</em>{23}} \right)$

Example 2.4

• Four charges at the corners of a square $ABCD$ of side $d$.

  • (a) Work required to put together the arrangement.

  • (b) Extra work needed to bring $q_0$ to the center $E$.

Solution:

• (a) Work = $\frac{q^2}{4 \pi \epsilon_0} \left( \frac{-4}{d} + \frac{2}{\sqrt{2}d} \right)$

• (b) The potential at center is zero. So, no extra work is needed.

Potential Energy in an External Field

• What is the potential energy of a charge $q$ in a given field?

• The external field $E$ is not produced by the given charge(s).

• The external electric field $E$ and corresponding external potential $V$ may vary from point to point.

• $V$ at a point $P$ is the work done in bringing a unit positive charge from infinity to $P$.

• Work done in bringing a charge $q$ from infinity to $P$ is $qV$.

• Potential energy of $q$ at $r$ in an external field = $qV(r)$.

• If an electron with charge $q = e = 1.6 \times 10^{-19} C$ is accelerated by a potential difference of $\Delta V = 1$ volt, it gains energy of $q \Delta V = 1.6 \times 10^{-19}J$.

• This unit of energy is defined as 1 electron volt or 1eV, i.e., $1 eV = 1.6 \times 10^{-19}J$.

Potential Energy of a System of Two Charges

• Potential energy of a system of two charges $q<em>1$ and $q</em>2$ located at $r<em>1$ and $r</em>2$ respectively, is:
  $U = q<em>1 V(r</em>1) + q<em>2 V(r</em>2) + \frac{1}{4 \pi \epsilon<em>0} \frac{q</em>1 q<em>2}{r</em>{12}}$

• $r<em>{12}$ is the distance between $q</em>1$ and $q_2$.

Example 2.5

• (a) Electrostatic potential energy of a system of two charges $7 \mu C$ and $-2 \mu C$ at (–9 cm, 0, 0) and (9 cm, 0, 0).

• (b) Work required to separate them infinitely far apart.

• (c) Electrostatic energy of the same in an external field $E = A (1/r^2), A = 9 \times 10^5 NC^{-1} m^2$

Solution:

• (a) $-0.7 J$

• (b) $0.7J$

• (c) The net electrostatic energy is $49.3 J$

Potential Energy of a Dipole in an External Field

• Consider a dipole with charges $q<em>1 = +q$ and $q</em>2 = –q$ in a uniform electric field $E$.

• It experiences no net force, but experiences a torque:
  $\tau = p \times E$

• The torque rotates it (unless $p$ is parallel or antiparallel to $E$).

Potential Energy of the Dipole

• External torque $\tau<em>{ext}$ rotates the dipole from angle $\theta</em>0$ to angle $\theta$:
  $W = \int^{\theta}<em>{\theta</em>0} \tau<em>{ext} d \theta = pE (\cos \theta</em>0 - \cos \theta)$

• With $\theta_0 = \pi/2$:
  $U(\theta) = -pE \cos \theta = -p.E$

Eq. (2.29) to Dipole

• $U = q [V(r<em>1) – V (r</em>2)] + \frac{1}{4 \pi \epsilon_0} \frac{q}{2a \cos \theta}$

• $[V(r1)–V (r2)] = –E \times 2a \cos \theta$.</p></li><li><p>Then$U = -p.E$</p></li></ul><h5 id="df441076-399f-4ebf-be53-224ebbcb0ece" data-toc-id="df441076-399f-4ebf-be53-224ebbcb0ece" collapsed="false" seolevelmigrated="true">Example 2.6</h5><ul><li><p>Molecule has permanent dipole moment$10^{-29} Cm$. A mole polarized by a strong field of$10^6 Vm^{-1}$. Field is suddenly changed by$60^\circ$. Heat released?</p></li></ul><h6 id="396f8e76-2432-4a41-b8e4-3837f993c985" data-toc-id="396f8e76-2432-4a41-b8e4-3837f993c985" collapsed="false" seolevelmigrated="true">Solution:</h6><p>Heat released =$3J$</p><h4 id="a260f9c7-e39b-42b8-8ef2-977026dd17e3" data-toc-id="a260f9c7-e39b-42b8-8ef2-977026dd17e3" collapsed="false" seolevelmigrated="true">Electrostatics of Conductors</h4><ul><li><p>Conductors contain mobile charge carriers.</p></li><li><p>In metallic conductors, charge carriers are electrons.</p></li></ul><h4 id="95cd35b7-c5e8-4ab5-a639-d9bdaca73533" data-toc-id="95cd35b7-c5e8-4ab5-a639-d9bdaca73533" collapsed="false" seolevelmigrated="true">Important Results Regarding Electrostatics of Conductors</h4><ul><li><p>1. Inside a conductor, the electrostatic field is zero.</p></li><li><p>2. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.</p></li><li><p>3. The interior of a conductor can have no excess charge in the static situation.</p></li><li><p>4. Electrostatic potential is constant throughout the volume of the conductor and has the same value on its surface.</p></li><li><p>5. Electric field at the surface of a charged conductor is$\vec{E} = \frac{\sigma}{\epsilon_0} \hat{n}$</p></li><li><p>where$\sigma$is the surface charge density and$\hat{n}$is a unit vector normal to the surface in the outward direction.</p></li><li><p>6. Electrostatic shielding</p></li></ul><h5 id="ed4a6bea-1f5d-4a0a-b1a3-601db3d2676a" data-toc-id="ed4a6bea-1f5d-4a0a-b1a3-601db3d2676a" collapsed="false" seolevelmigrated="true">Example 2.7</h5><ul><li><p>(a) A comb rubbed through dry hair attracts bits of paper. Why?</p></li><li><p>(b) Ordinary rubber is an insulator. Special tyres are slightly conductive. Why?</p></li><li><p>(c) Vehicles carrying flammable materials have ropes touching the ground. Why?</p></li><li><p>(d) A bird perches on a high power line, and nothing happens. A man touching the line gets a shock. Why?</p></li></ul><h6 id="2e26afae-3be0-439d-a68e-1a78cd195e64" data-toc-id="2e26afae-3be0-439d-a68e-1a78cd195e64" collapsed="false" seolevelmigrated="true">Solution:</h6><ul><li><p>(a)Comb is charged and polarizes molecules in paper.</p></li><li><p>(b)Conduct charge to the ground.</p></li><li><p>(c)Similar to (b).</p></li><li><p>(d)Potential difference.</p></li></ul><h4 id="dd37b2d4-0cb9-4a37-a0b5-e84960fb024f" data-toc-id="dd37b2d4-0cb9-4a37-a0b5-e84960fb024f" collapsed="false" seolevelmigrated="true">Dielectrics and Polarization</h4><ul><li><p>Dielectrics are non-conducting substances.</p></li></ul><h4 id="1c9294c2-404d-48e9-9131-d7e8cbf68090" data-toc-id="1c9294c2-404d-48e9-9131-d7e8cbf68090" collapsed="false" seolevelmigrated="true">Polar and Non-Polar Molecules</h4><ul><li><p>In a non-polar molecule, the centers of positive and negative charges coincide.</p></li><li><p>Examples are Oxygen ($O2$) and Hydrogen ($H2$).</p></li><li><p>A polar molecule has separated centers of positive and negative charges.</p></li><li><p>Examples are HCl or Water ($H_2O$).</p></li></ul><h4 id="ad164a2f-d301-417b-a509-b9e2237a985e" data-toc-id="ad164a2f-d301-417b-a509-b9e2237a985e" collapsed="false" seolevelmigrated="true">Effect of External Electric Field</h4><ul><li><p>In an external electric field, non-polar molecules develop an induced dipole moment.</p></li><li><p>Polar molecules align with the field.</p></li><li><p>The dipole moment per unit volume is called Polarization ( P ).</p></li><li><p>For linear isotropic dielectrics,$P = \epsilon0 \chie E$</p></li></ul><h4 id="2d3462e2-ce37-4228-9873-ca013ed43ac8" data-toc-id="2d3462e2-ce37-4228-9873-ca013ed43ac8" collapsed="false" seolevelmigrated="true">Surface Charges</h4><ul><li><p>At the surfaces of the dielectric normal to the electric field, there is a net charge density ($\pm \sigma_p$).<br>*(arises from bound charges)</p></li><li><p>The total field is thereby reduced.</p></li></ul><h4 id="c3be56f6-78cc-4db9-b3fc-28066514045d" data-toc-id="c3be56f6-78cc-4db9-b3fc-28066514045d" collapsed="false" seolevelmigrated="true">Capacitors and Capacitance</h4><ul><li><p>A capacitor is a system of two conductors separated by an insulator.</p></li><li><p>The relation between charge and potential is:$Q = CV$.</p></li><li><p>C is called the capacitance of the capacitor.</p></li><li><p>The SI unit of capacitance is 1 farad (=1 coulomb volt-1) or 1 F = 1 C V–1.</p></li><li><p>The maximum electric field that a dielectric medium can withstand without break-down (of its insulating property) is called its dielectric strength; for air it is about$3 \times 10^6 Vm^{-1}$.</p></li></ul><h4 id="427dac5f-58a7-4940-8186-c259575acaf6" data-toc-id="427dac5f-58a7-4940-8186-c259575acaf6" collapsed="false" seolevelmigrated="true">Parallel Plate Capacitor</h4><ul><li><p>Parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.</p></li><li><p>Using Eq. (1.33), the electric field in different regions is:</p></li><li><p>Outer region I.$E = 0$</p></li><li><p>Outer region II.$E = 0$</p></li><li><p>In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving<br>$E = \frac{\sigma}{\epsilon_0}$.</p></li><li><p>The direction of the electric field is from the positive to the negative plate.</p></li><li><p>Now for a uniform electric field, the potential difference is:<br>$V = Ed = \frac{1}{\epsilon_0} \frac{Qd}{A}$</p></li><li><p>The capacitance C of the parallel plate capacitor is then:<br>$C = \frac{Q}{V}$<br>$C = \frac{\epsilon_0 A}{d}$</p></li></ul><h4 id="29202dfb-41e7-4e04-9ca4-e173276f06ed" data-toc-id="29202dfb-41e7-4e04-9ca4-e173276f06ed" collapsed="false" seolevelmigrated="true">Effect of Dielectric on Capacitance</h4><ul><li><p>Suppose that the medium between the plates is fully occupied by the intervening region.</p></li><li><p>The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is$±( \sigma - \sigma_P )$.</p></li><li><p>Dielectric constant is$K =\frac{ \sigma }{ \sigma - \sigma_P }$</p></li><li><p>With dielectric:<br>$V = \frac{Qd}{KA \epsilon0 }$$C = \frac{K \epsilon0 A}{d}$</p></li><li><p>$ \epsilon = K \epsilon_0$(permittivity of the medium</p></li><li><p>$K$(dielectric constant)</p></li></ul><h5 id="7e62a2e6-cb53-4989-af18-deeb66899975" data-toc-id="7e62a2e6-cb53-4989-af18-deeb66899975" collapsed="false" seolevelmigrated="true">Example 2.8</h5><ul><li><p>A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor.</p></li><li><p>It has thickness$(3/4)d$, where$d$is the separation of the plates</p></li><li><p>How is the capacitance changed when the slab is inserted between the plates?</p></li></ul><h6 id="7156fc12-8581-4861-b647-b45114e17de2" data-toc-id="7156fc12-8581-4861-b647-b45114e17de2" collapsed="false" seolevelmigrated="true">Solution:</h6><ul><li><p>Capacitance increases$ C = \frac{4C_0 K}{K+3}$</p></li></ul><h4 id="6e451657-c136-4d7a-8793-13a36131cb88" data-toc-id="6e451657-c136-4d7a-8793-13a36131cb88" collapsed="false" seolevelmigrated="true">Combination of Capacitors - Series</h4><ul><li><p>Charges on all plates are same. Q</p></li><li><p>V = V1 + V2, with V1 = Q/C1 and V2 = Q/C2</p></li><li><p>$C = \frac{(C1 C2 )}{C1 + C2}$</p></li><li><p>In general, 1/C = 1/C1 + 1/C2 + .. + 1/Cn</p></li></ul><h4 id="0badc5d0-47a3-441e-b878-38a454c7dd27" data-toc-id="0badc5d0-47a3-441e-b878-38a454c7dd27" collapsed="false" seolevelmigrated="true">Combination of Capacitors - Parallel</h4><ul><li><p>The potential difference is the same: V</p></li><li><p>The total charge is now Q = Q1 + Q2</p></li><li><p>Hence C = C1 + C2</p></li><li><p>In general, C = C1 + C2 + … + Cn</p></li></ul><h5 id="aba8ee64-4f2b-4514-9686-080624d9576b" data-toc-id="aba8ee64-4f2b-4514-9686-080624d9576b" collapsed="false" seolevelmigrated="true">Example 2.9</h5><p>A network of four 10 \mu F capacitors is connected to a 500 V supplyDetermine (a) the equivalent capacitance of the network and (b) the charge on each capacitor</p><h6 id="b032cb55-8242-4293-be4e-7a7b81fbcef0" data-toc-id="b032cb55-8242-4293-be4e-7a7b81fbcef0" collapsed="false" seolevelmigrated="true">Answer</h6><p>(a) The equivalent capacitance is C =13.3\mu F. (b)The charge on each of the capacitors,C1, C2 and C3 is Q = 1.7 \times 10^{-3}C, and Q’ = 5.0 \times 10^{-3} C</p><h4 id="14e39b10-85dd-472e-b740-24d83f8bfd2e" data-toc-id="14e39b10-85dd-472e-b740-24d83f8bfd2e" collapsed="false" seolevelmigrated="true">Energy Stored In a Capacitor</h4><ul><li><p>Consider two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q at the end .</p></li><li><p>Work is stored in potential energy U =$\frac{1}{2} \frac{Q^2}{C} $=$\frac{1}{2} CV^2 $=$\frac{1}{2} QV$$
  Since electrostatic force is conservative, this work is stored in the form of potential energy of the system. For the same reason, the final result for potential energy
  *
  Energy density u =(1/2)0E2