XI-Physics CH-16

Electrostatics

TOPIC 1Electric Potential Difference
SQ 16.1.1

Define electric potential difference.
Definition
The electric potential difference is the work done per unit charge in moving it between two points against the electric field.
Formula
$$\Delta V=V_B-V_A=\frac{W_{AB}}{q_0}=\frac{\Delta U}{q_0}$$
SQ 16.1.2

Write the SI unit of potential difference.
SI Unit
Its SI unit is the volt, where
$$1\,\mathrm{V}=\frac{1\,\mathrm{J}}{1\,\mathrm{C}}.$$
It can be converted into other units such as kilovolts or millivolts.
SQ 16.1.3

Write the relation between potential difference and electric field.
Formula
$$\Delta V=-\mathbf{E}\cdot\Delta\mathbf{d}$$
SQ 16.1.4

Define potential gradient and write the field in terms of it.
Definition
The electric field represents the rate of change of potential with distance, which is called the gradient of potential.
Formula
$$E=-\frac{\Delta V}{\Delta d}$$
SQ 16.1.5

What does the negative sign in E = −ΔV/Δd indicate?
Answer
The negative sign indicates that the field points from higher to lower potential.
SQ 16.1.6

Write the relation for a uniform electric field.
Formula
$$E=\frac{V}{d}$$
This shows that the potential difference between two points is directly proportional to the electric field strength and the distance between them.
SQ 16.1.7

What does a stronger electric field cause?
Answer
A stronger electric field causes a more rapid decrease in potential, providing a clear physical link between field intensity and voltage variation in space.
SQ 16.1.8

Why is the Earth assumed to be at zero potential?
Answer
Although points at infinity are assumed to be at zero potential, practically the Earth is assumed to be at zero potential. Therefore the absolute value of electric potential at a point is measured with respect to the Earth.
SQ 16.1.9

A 6 V battery is connected to plates 0.3 cm apart. Find the field intensity between them.
Solution
Using
$$E=\dfrac{\Delta V}{\Delta d}$$
Calculation
$$E=\frac{6}{0.003}$$
Result
$$E=2000\,\mathrm{V\,m^{-1}}$$
TOPIC 2Electric Potential
SQ 16.2.1

Define electric potential at a point.
Definition
The electric potential at any point in an electric field is equal to the work done in bringing a unit positive charge from infinity to that point, keeping it in electrostatic equilibrium.
Formula
$$V=\frac{W}{q_0}$$
SQ 16.2.2

Why are potential and potential difference scalar quantities?
Reason
Both potential and potential difference are scalar quantities because both the work $W$ and the charge $q_0$ are scalars.
SQ 16.2.3

Is the potential at a point really a potential difference?
Answer
Yes. The potential at a point is still the potential difference between the potential at that point and the potential at infinity.
SQ 16.2.4

Why can ΔV = −EΔr not be used directly for a point charge?
Reason
The relation requires the electric intensity $E$ to remain constant. In the field of a point charge, $E$ varies inversely as the square of the distance, so it no longer remains constant.
SQ 16.2.5

How is the difficulty of a varying field overcome in deriving potential?
Method
Two points $A$ and $B$ are taken infinitesimally close to each other, so that $E$ remains almost constant between them.
SQ 16.2.6

Which approximation is used for the mid-point distance in the derivation?
Approximation
As the points $A$ and $B$ are very close, as a first approximation the arithmetic mean is taken to be equal to the geometric mean, which gives
$$r^{2}=r_Ar_B$$
SQ 16.2.7

Write the potential difference between two nearby points in a radial field.
Formula
$$V_A-V_B=\frac{1}{4\pi\varepsilon_0}q\left(\frac{1}{r_A}-\frac{1}{r_B}\right)$$
SQ 16.2.8

Derive the absolute electric potential due to a point charge.
Derivation
To calculate the absolute potential at $A$, the point $B$ is assumed to be at infinity so that
$$V_B=0$$
and
$$\dfrac{1}{r_B}=0$$
Result
$$V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}$$
SQ 16.2.9

When is the electric potential negative?
Answer
The electric potential will be negative if the charge $q$ is negative.
SQ 16.2.10

How is the total potential found when there are several charges?
Answer
Each charge contributes to the total electric potential. At any point the total electric potential is the algebraic sum of the individual potentials created by all the charges.
SQ 16.2.11

Find the electric potential 1.2 m from a charge of 5.0 × 10⁻⁸ C.
Solution
Using
$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$$
Calculation
$$V=\frac{9\times10^{9}\times5.0\times10^{-8}}{1.2}$$
Result
$$V=375\,\mathrm{V}$$
SQ 16.2.12

For a pair of unequal opposite charges, where are the zero potential points?
Answer
The potential is zero where the potential due to the positive charge is exactly equal in magnitude to that due to the negative charge.
Location
Both points lie closer to the smaller charge, one between the two charges and the other outside the negative charge.
TOPIC 3Electric Potential Energy
SQ 16.3.1

How does a positive charge move freely in an electric field?
Answer
If a positive charge is allowed to move freely in an electric field, it will always accelerate from a point of higher potential to a point of lower potential.
SQ 16.3.2

What happens to the potential energy of a freely moving charge?
Answer
The potential energy of the charge decreases by
$$\Delta U=q_0\Delta V$$
This decrease in potential energy appears as its kinetic energy.
SQ 16.3.3

Write the velocity gained by a charge falling through a potential difference.
Formula
$$q_0\Delta V=\frac{1}{2}mv^{2}$$
SQ 16.3.4

How does a negative charge move in an electric field?
Answer
A negative charge allowed to move freely in an electric field moves from a point of lower potential to a point of higher potential.
SQ 16.3.5

Define the electron volt and give its value.
Definition
The electron volt is the energy acquired or lost by an electron which has been accelerated through a potential difference of $1$ volt.
Value
$$1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}$$
SQ 16.3.6

Derive the electric potential energy of two point charges.
Derivation
A charge $Q$ at point $A$ creates a potential $V$ at point $B$. If a charge $q$ is brought from infinity to $B$, the work done is
$$W=qV$$
Result
$$U=\frac{1}{4\pi\varepsilon_0}\frac{Qq}{r}$$
SQ 16.3.7

What is stored as electric potential energy in a two-charge system?
Answer
By definition, the work done in placing the two charges at points $A$ and $B$ is stored as electric potential energy in the system.
SQ 16.3.8

Two charges of 8 μC and 5 μC are 4.0 cm apart. Find their potential energy.
Solution
Using
$$U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{r}$$
Calculation
$$U=\frac{9\times10^{9}\times8\times10^{-6}\times5\times10^{-6}}{4\times10^{-2}}$$
Result
$$U=9\,\mathrm{J}$$
TOPIC 4Capacitors
SQ 16.4.1

Define a capacitor.
Definition
A capacitor is a device that can store electrical energy in the form of an electric field.
SQ 16.4.2

Name the two types of capacitors discussed in the book.
Types
Isolated spherical conductors and parallel plate capacitors.
SQ 16.4.3

Define an isolated spherical conductor.
Definition
A spherical conductor placed in space such that it is far from other conductors or influence is known as an isolated spherical conductor.
SQ 16.4.4

Define capacitance of an isolated spherical conductor.
Definition
The capacitance of an isolated hollow spherical conductor is the electric charge stored on the sphere per unit electric potential at its surface.
Formula
$$C=\frac{Q}{V}$$
SQ 16.4.5

Define one farad for a spherical conductor.
Definition
For a spherical conductor, $1$ farad is the capacitance of a sphere that stores $1$ coulomb of charge when its electric potential is $1$ volt, assuming it is isolated in free space.
SQ 16.4.6

Derive the capacitance of an isolated spherical conductor in vacuum.
Derivation
Since
$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$$
substituting in
$$C=\dfrac{Q}{V}$$
gives
Result
$$C_{vac}=4\pi\varepsilon_0R$$
SQ 16.4.7

Write the capacitance of a spherical conductor in a medium.
Formula
$$C_{med}=4\pi\varepsilon_0\varepsilon_rR$$
Here $\varepsilon_r$ is the dielectric constant of the medium.
SQ 16.4.8

On what does the capacitance of a spherical conductor depend?
Answer
The capacitance of a spherical conductor is independent of the charge or potential. It depends only on the radius of the sphere and the medium.
SQ 16.4.9

What is the effect of increasing the radius of a spherical conductor?
Answer
The capacitance increases with an increase in the radius of the sphere. This means larger spheres store more charge at the same surface potential.
TOPIC 5Parallel Plate Capacitors
SQ 16.5.1

Describe a parallel plate capacitor.
Description
A parallel plate capacitor consists of two conductors placed near one another, separated by vacuum, air or any other insulator known as a dielectric. The conductors are usually in the form of parallel plates.
SQ 16.5.2

What happens when a parallel plate capacitor is connected to a battery?
Answer
It establishes a potential difference of $V$ volts between the two plates. The battery places a charge $+Q$ on the plate connected to its positive terminal and $-Q$ on the other plate.
SQ 16.5.3

Define the capacitance of a capacitor.
Definition
The capacitance of a capacitor is the amount of charge on one plate necessary to raise the potential of that plate by one volt with respect to the other.
Formula
$$C=\frac{Q}{V}$$
SQ 16.5.4

What does capacitance measure?
Answer
Capacitance depends upon the geometry of the plates and the medium between them. It is a measure of the ability of a capacitor to store charge.
SQ 16.5.5

Why must the separation between the plates be small?
Reason
The distance $d$ is kept small so that the electric field between the plates is uniform and confined almost entirely in the region between the plates.
SQ 16.5.6

Derive the capacitance of a parallel plate capacitor in vacuum.
Derivation
Since
$$E=\dfrac{V}{d}$$
and by Gauss’s law
$$E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0A}$$
Equating and substituting in
$$C=\dfrac{Q}{V}$$
Result
$$C=\frac{\varepsilon_0A}{d}$$
SQ 16.5.7

Write the capacitance of a parallel plate capacitor with a dielectric.
Formula
$$C=\frac{\varepsilon_0\varepsilon_rA}{d}$$
The capacitance is enhanced by the factor $\varepsilon_r$.
SQ 16.5.8

Describe the experiment showing the effect of a dielectric.
Experiment
A charged capacitor has its plates connected to a voltmeter. When a dielectric material is inserted between the plates, the reading drops, indicating a decrease in the potential difference.
Conclusion
Since
$$C=\dfrac{Q}{V}$$
and $V$ decreases while $Q$ remains constant, the value of $C$ increases.
SQ 16.5.9

Define dielectric constant.
Definition
The dielectric constant is the ratio of the capacitance of a parallel plate capacitor with an insulating substance as medium between the plates, to its capacitance with vacuum or air as medium.
Formula
$$\varepsilon_r=\frac{C_{med}}{C_{vac}}$$
SQ 16.5.10

What is polarization of a dielectric?
Definition
A dielectric is an insulator with no free electrons. When such an atom is subjected to an electric field, its electrons are displaced slightly relative to the nucleus in a direction opposite to the field. This phenomenon is known as polarization, and the distorted atom is said to be polarized.
SQ 16.5.11

Why does the potential difference decrease when a dielectric is inserted?
Reason
The atoms of the dielectric get polarized, so charges appear on those surfaces of the slab in contact with the metal plates. The polarity of these charges is opposite to that of the adjacent plate.
Result
This effectively decreases the charge on the plates, so the electric intensity and the potential difference decrease.
SQ 16.5.12

How does the capacitance change if the separation between the plates is increased?
Answer
Since
$$C=\dfrac{\varepsilon_0A}{d}$$
the capacitance is inversely proportional to the separation. Therefore increasing the separation decreases the capacitance.
TOPIC 6Combined Capacitance of Capacitors
SQ 16.6.1

What is a combination of capacitors and why is it made?
Definition
When two or more capacitors are connected together in a circuit, it is known as a combination of capacitors.
Purpose
It is done to obtain a desired capacitance, and multiple connections behave as a single equivalent capacitor.
SQ 16.6.2

Name the two common types of combination of capacitors.
Types
Parallel combination and series combination.
SQ 16.6.3

Describe the parallel combination of capacitors.
Description
The plates of one side of all the capacitors are connected to the negative terminal and the plates of the other side to the positive terminal of the battery.
Property
The potential difference across each capacitor is the same, while the charge on each will be different depending upon its capacitance.
SQ 16.6.4

Derive the equivalent capacitance for a parallel combination.
Derivation
Since the total charge is
$$Q=Q_1+Q_2+Q_3$$
and each capacitor has the same potential difference $V$, therefore
$$C_eV=C_1V+C_2V+C_3V.$$
Result
$$C_e=C_1+C_2+C_3$$
SQ 16.6.5

State the rule for capacitors connected in parallel.
Rule
When capacitors are connected in parallel, their combined capacitance is equal to the sum of their individual capacitances.
SQ 16.6.6

Describe the series combination of capacitors.
Description
The capacitors are connected one after the other in a line, with the second plate of one capacitor connected to the first plate of the next.
SQ 16.6.7

Why does each capacitor in series carry the same charge?
Reason
The battery sends charge $+Q$ to the first plate of $C_1$. Since the second plate is insulated it cannot draw charge from the battery, but the positive charge on the first plate induces an equal negative charge on it.
Result
Thus each capacitor receives a charge of magnitude $Q$ on each of its plates.
SQ 16.6.8

Write the potential relation for a series combination.
Formula
$$V=V_1+V_2+V_3$$
The potential difference across each capacitor is different, depending upon its capacitance.
SQ 16.6.9

Write the equivalent capacitance for a series combination.
Formula
$$\frac{1}{C_e}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$
SQ 16.6.10

An 80 pF capacitor charged to 48 V is connected in parallel to another, dropping to 30 V. Find the charge shifted.
Solution
Initial charge
$$Q_1=80\times10^{-12}\times48=3840\times10^{-12}\,\mathrm{C}$$
and the new charge on $C_1$ is $2400\times10^{-12}\,\mathrm{C}$.
Result
$$Q_2=1440\times10^{-12}\,\mathrm{C}$$
SQ 16.6.11

For the above problem, find the capacitance of the second capacitor.
Solution
$$C_2=\frac{Q_2}{V}=\frac{1440\times10^{-12}}{30}$$
Result
$$C_2=48\,\mathrm{pF}$$
SQ 16.6.12

Find the equivalent capacitance of 4 μF and 8 μF connected in series.
Solution
$$\frac{1}{C_e}=\frac{1}{4}+\frac{1}{8}$$
Result
$$C_e=2.67\,\mu\mathrm{F}$$
TOPIC 7Energy Stored in a Capacitor
SQ 16.7.1

How is a capacitor also a device for storing energy?
Answer
A capacitor is a device to store charge. Alternatively, it is possible to think of a capacitor as a device for storing electrical energy.
SQ 16.7.2

Give evidence of energy stored in a charged capacitor.
Evidence
If we connect the two plates of a charged capacitor by a thick wire, the charge flows from one plate to the other and the wire becomes hot.
Conclusion
Since heat is a form of energy, the capacitor possessed energy when it was charged.
SQ 16.7.3

Why is the graph of V against Q a straight line?
Reason
The charge $Q$ on the plates increases with the increase in potential $V$, that is $Q\propto V$. Therefore the graph is a straight line passing through the origin.
SQ 16.7.4

How is the energy stored found graphically?
Method
The work required to move a small charge $\Delta Q$ through a potential difference $V$ is
$$\Delta W=(\Delta Q)V$$
which is the area of a thin rectangle under the graph.
Result
The total work is the area of the triangle under the $V$ against $Q$ graph.
SQ 16.7.5

Write the three expressions for energy stored in a capacitor.
Formulas
$$W=\frac{1}{2}QV$$
$$W=\frac{1}{2}CV^{2}$$
$$W=\frac{Q^{2}}{2C}$$
SQ 16.7.6

Where is the energy of a charged capacitor stored?
Answer
Energy can be considered to be stored in the electric field between the plates, rather than as the potential energy of the charges on the plates.
SQ 16.7.7

Define energy density of a capacitor.
Definition
Energy density is the energy stored per unit volume between the plates.
Formula
$$\text{Energy density}=\frac{\text{Energy}}{\text{Volume}}$$
This equation is valid for any electric field strength.
SQ 16.7.8

Two capacitors of 4 μF and 8 μF in series are charged by 120 V. Find the total charge.
Solution
The equivalent capacitance is $2.67\,\mu\mathrm{F}$, so
$$Q=C_eV$$
Calculation
$$Q=2.67\times10^{-6}\times120$$
Result
$$Q=3.2\times10^{-4}\,\mathrm{C}$$
SQ 16.7.9

A 16 μF capacitor is charged to 100 V. Find the energy stored.
Solution
$$W=\frac{1}{2}CV^{2}=\frac{1}{2}(16\times10^{-6})(100)^{2}$$
Result
$$W=0.08\,\mathrm{J}$$
SQ 16.7.10

By what factor does the energy increase if the potential difference is doubled?
Answer
Since the energy depends on $V^{2}$, if $V$ is doubled the energy increases by a factor of 4.
SQ 16.7.11

How does a defibrillator use a capacitor?
Answer
A device called a defibrillator uses the charge stored in a capacitor to deliver a controlled electric shock that can restore normal heart rhythm.
TOPIC 8Charging and Discharging a Capacitor
SQ 16.8.1

What is an RC circuit?
Definition
Many electric circuits consist of both capacitors and resistors. Such a resistor-capacitor circuit is called an RC circuit.
SQ 16.8.2

How does a capacitor charge in an RC circuit?
Answer
The capacitor is not charged immediately; rather the charge builds up gradually to the equilibrium value
$$q_0=CV_0$$
SQ 16.8.3

How is the voltage across a charging capacitor obtained?
Answer
The voltage across the capacitor at any instant is obtained by dividing the charge by the capacitance, that is
$$V=\dfrac{q}{C}$$
SQ 16.8.4

On what does the rate of charging or discharging depend?
Answer
How fast or how slow the capacitor charges or discharges depends upon the product of the resistance and the capacitance used in the circuit.
SQ 16.8.5

Define the time constant of an RC circuit.
Definition
Since the unit of the product $RC$ is that of time, this product is known as the time constant. It is the time required by the capacitor to deposit $0.63$ times the equilibrium charge.
SQ 16.8.6

What is the effect of a small time constant on charging?
Answer
The charge reaches its equilibrium value sooner when the time constant is small.
SQ 16.8.7

Describe the discharging of a capacitor through a resistor.
Answer
When the switch is set so that the capacitor is connected across the resistor, the charge on the left plate flows through the resistance and neutralizes the charge on the right plate.
Graph
Discharging begins at
$$t=0$$
when
$$q=CV_0$$
and decreases gradually to zero.
SQ 16.8.8

What is the effect of the time constant on discharging?
Answer
Smaller values of the time constant $RC$ lead to a more rapid discharge.
SQ 16.8.9

Verify that an ohm times a farad is equivalent to a second.
Proof
From Ohm’s law
$$V=IR$$
and putting
$$I=\dfrac{q}{t}$$
gives
$$V=\dfrac{qR}{t}$$
so
$$t=\dfrac{qR}{V}$$
Since
$$C=\dfrac{q}{V}$$
therefore
$$t=RC$$
so the product of ohm and farad is the second.
TOPIC 9Exponential Decay of Discharging Capacitor
SQ 16.9.1

What does exponential mean?
Definition
Exponential means a process whose rate of change at any instant is proportional to the present value of the quantity, so it changes quickly at first and then more slowly.
SQ 16.9.2

Write the general exponential decay formula for a discharging capacitor.
Formula
$$x=x_0e^{-t/RC}$$
Here $x_0$ is the initial value of the quantity at
$$t=0$$
and $RC$ is the time constant.
SQ 16.9.3

What is Euler’s constant and what is its value?
Answer
In the decay equation, $e$ is a mathematical constant called Euler’s constant, which is also used as the base of the natural logarithm. Its value is approximately $2.718$.
SQ 16.9.4

What does the negative exponent in the decay formula indicate?
Answer
The negative exponent indicates that the quantity decreases exponentially over time.
SQ 16.9.5

Write the decay equations for charge, voltage and current.
Formulas
$$Q=Q_0e^{-t/RC}$$
$$V=V_0e^{-t/RC}$$
$$I=I_0e^{-t/RC}$$
SQ 16.9.6

Describe the shape of the decay graph.
Answer
The graph starts with a steep slope, indicating fast discharge, and gradually flattens out as time progresses, indicating a slower discharge rate.
SQ 16.9.7

Give the water tank analogy for the decay of charge.
Analogy
The water level and pressure are analogous to charge and voltage, and the flow rate of water is analogous to the current.
Behaviour
When the level is high the flow is fast, and as the level comes down the rate of flow slows down due to the decreasing pressure.
SQ 16.9.8

Define the time constant in terms of exponential decay.
Derivation
Putting
$$t=RC$$
in
$$Q=Q_0e^{-t/RC}$$
gives
$$Q=Q_0e^{-1}$$
Definition
The time constant is the duration of time in which the initial charge $Q_0$ drops to about $0.37Q_0$, that is $37\%$ of its initial value.
SQ 16.9.9

What is the effect of a large or small time constant on circuit response?
Answer
A smaller time constant means the circuit responds faster, while a larger time constant means it responds more slowly.
SQ 16.9.10

Write the half-life of the charge on a discharging capacitor.
Derivation
Putting
$$Q=\dfrac{1}{2}Q_0$$
in the decay equation and using natural logarithms.
Result
$$t_{1/2}=0.69\,RC$$
SQ 16.9.11

Which other physical phenomenon is described by exponential functions?
Answer
Exponential functions frequently describe phenomena in physics such as radioactive decay, in which the rate of change in a process depends directly on its current value.
TOPIC 10Use of Capacitors
SQ 16.10.1

Why are capacitors used in refrigerators and air-conditioners?
Reason
The compressors typically use single phase motors which can struggle to start on their own. A capacitor of high capacitance stores a big amount of electrical energy and releases it to the motor winding of the compressor.
SQ 16.10.2

What is a start capacitor?
Definition
A start capacitor acts like a “jump start” for the motor, providing a boost of power to get the motor rotating. Once the motor reaches a certain speed, a relay disconnects the capacitor as it is no longer needed.
SQ 16.10.3

What is a run capacitor?
Definition
Most refrigerators and air-conditioners also use a run capacitor which stays connected in the circuit to improve energy efficiency.
SQ 16.10.4

How do capacitors form a time delay circuit?
Answer
Capacitors in conjunction with resistors can form a time delay circuit. It is widely used in refrigerators and air conditioners to create a delay in triggering the power supply, as a precautionary measure to protect the devices from power fluctuation.
SQ 16.10.5

Explain the working of a capacitor in a flashgun.
Working
The capacitor accumulates charge from the battery of the camera. When the shutter button is pressed, the capacitor is connected to the flash tube.
Result
The rapid discharge provides the high current needed to ionize the xenon gas inside the tube, causing it to emit a burst of bright light.
SQ 16.10.6

How are capacitors used as filters in rectifier circuits?
Answer
Capacitors are widely used in rectifier circuits for the smoothing of pulsating DC into smooth DC.
SQ 16.10.7

Give the medical uses of capacitors.
Uses
Capacitors store the electrical energy needed to deliver a life-saving shock to the heart in case of cardiac arrest.
They are also used in imaging machines such as CT, MRI, X-ray and ultrasound, helping to filter out noise and improve image clarity, and they filter unwanted signals in EEG and ECG machines.