Electrostatics - 203PHYS

203 PHYS - Physics 2203 PHYS - Physics 2
(for Engineering)(for Engineering)
Instructor:
Dr. Sabar Hutagalung
Physics Department, Faculty of Science, Jazan University
Jazan, Saudi Arabia
Email: sdhutagalung@gmail.com

Main ReferenceMain Reference
• Raymond A. Serway & John W. Jewett, Jr., Physics for
Scientists and Engineers with Modern Physics, 9th Edition,
Brooks/Cole, 2014.

Electrostatics
• Coulomb’s LawCoulomb’s Law
• Electric FieldsElectric Fields
• Gauss’s Law and Electric FluxGauss’s Law and Electric Flux
• Electric PotentialElectric Potential
• CapacitorsCapacitors

Charge and Mass
Electric chargeElectric charge is a basic property of matter.
Two basic charges:
Negative (electron) and positive (proton)

Law of Electric Charges
AttractedAttracted
RepelledRepelled
Experiments have shown that
Like signed charges repel each other: (+ +) or (- -)Like signed charges repel each other: (+ +) or (- -)
Unlike signed charges attract each other: (+ -)Unlike signed charges attract each other: (+ -)

The electric force between two point charges.The electric force between two point charges.
2
21
r
qq
kF ee =
ke = Coulomb’s constant = 8.9876 x 109
N.m2
/C2
q1 = charge on mass 1; q2 = charge on mass 2
r = the distance between the two charges
• Coulomb’s law describes the interaction between
bodies due to their charges.
InverseInverse
SquareSquare
LawLaw
Coulomb’s LawCoulomb’s Law

• For practical reasons, the coulomb is defined
using current and magnetism giving
ke ≅ 9 x 109
Nm2
/C2
• Permittivity of free space
2212
0 Nm/C1084.8
4
1 −
×==
ekπ
ε
Then
2
21
04
1
r
qq
Fe
πε
=
Coulomb’s LawCoulomb’s Law

Example
• The electron and proton of a hydrogen atom are
separated (on the average) by a distance of
approximately 5.3 x 10-11
m. Find the magnitudes
of the electric forcethe electric force and the gravitational forcethe gravitational force
between the two particles.

Answer
• Use Coulomb’s law:
• Use Newton’s law of universal gravitation
The ratioThe ratio FFee/F/Fgg = 2 x 10= 2 x 103939
. Therefore, the gravitational force. Therefore, the gravitational force
between atomic particles is negligible when compared with thebetween atomic particles is negligible when compared with the
electric force.electric force.
GG

Vector form of Coulomb’s lawVector form of Coulomb’s law
• Coulomb’s law expressed in vector form for the electric
force exerted by a charge q1 on a second charge q2,
written F12:
IfIf qq11 && qq22 have thehave the same signsame sign
as figure (a), the productas figure (a), the product qq11qq22 isis
positive and electric force ispositive and electric force is
repulsive.repulsive.
IfIf qq11 && qq22 are ofare of opposite signopposite sign
as figure (b), the productas figure (b), the product qq11qq22 isis
negative and electric force isnegative and electric force is
attractive.attractive.

Vector form of Coulomb’s lawVector form of Coulomb’s law
• When more than two chargesmore than two charges are present, the
resultant force on any one of them equals the
vector sum of the forces exerted by the other
individual charges.
• For example, if four charges are present, the
resultant force exerted by particles 2, 3, and 4 on
particle 1 is:

Solution
22√√qq22 -- xx√√qq22 = ±= ±xx√√qq11
xx√√qq22 ±± xx√√qq11 = 2= 2√√qq22

Analysis Model: Particle in a
Field (Electric)
• A small positive test charge placed at point near an object
carrying a much larger positive charge experiences an electric
field at point established by the source charge Q. We will
always assume that the test charge is so small that the field of
the source charge is unaffected by its presence.
• The electric field vector E at a point in space is defined as the
electric force Fe acting on a positive test charge q0 placed at
that point divided by the test charge:
Units: N/C

Particle in a Field (Electric)
• Electric force at a test charge q0 is placed at point P, a distance r
from the source charge:
• The electric field
at P created by q:

• The force Fo exerted on a point charge qo placed in an
electric field E.
The charge qo can be
either positive or
negative.
• If qqoo is positiveis positive, the
force Fo experienced by
the charge is the same
direction as E,
• if qqoo is negativeis negative, Fo and
E are in opposite
directions.

Electric field due to a finite
number of point charges
• The electric field at point P due to a group of source
charges can be expressed as the vector sum:
• where ri is the distance from the i-th source charge qi to
the point P and ri is a unit vector directed from qi toward
P.
Net field at P: E = E1 + E2 + E3

Quiz
• A test charge of +3 µC is at a point P where an
external electric field is directed to the right and
has a magnitude of 4 x 106
N/C. If the test charge
is replaced with another test charge of -3 µC,
what happens to the external electric field at P?
(a) It is unaffected. (b) It reverses direction. (c) It
changes in a way that cannot be determined.
Electric fieldElectric field EE direction depend on sign of sourcedirection depend on sign of source
chargecharge qq.. EE byby qq at pointat point PP is unaffected by sign ofis unaffected by sign of
test chargetest charge qqoo..

• Imagine an object with charge that we call a source
charge. The source charge establishes an electric field E
throughout space.
• Now imagine a particle with chargea particle with charge qq is placed in that
field. The particle interacts with the electric field so that
the particle experiences an electric force given by:

Example
• A water droplet of mass 3 x 10-12
kg is located in
the air near the ground during a stormy day. An
atmospheric electric field of magnitude 6 x 103
N/C points vertically downward in the vicinity of
the water droplet. The droplet remains
suspended at rest in the air. What is the electric
charge on the droplet?

Solution
• Newton’s 2nd law from the particle in equilibrium model in the
vertical direction:
• Using the two particle in a field models, substitute for the forces, the
vertical component of the electric field is negative:
• Solve for the charge on the water droplet:
Fe
Fg
E
The problem claims that the electric fieldelectric field is in the downward direction. The
electric forceelectric force is upward to balance the downward gravitational force.
Therefore, the charge is negativenegative so that the electric force is in the direction
opposite to the electric field.
mm = 3 x 10= 3 x 10-12-12
kgkg
EE = 6 x 10= 6 x 1033
N/C (N/C (downward)

Example
• Charges q1 and q2 are located on the x axis, at
distances a and b, respectively, from the origin as
shown in Figure 23.12.
(A) Find the components of the net electric field at
the point P, which is at position (0, y).
(B) Evaluate the electric field at point P in the
special case that q1=q2and a = b.
(C) Find the electric field due to the electric dipole
when point P is a distance y >> a from the origin.

Electric Flux
• Consider an electric field that is uniform in both magnitude
and direction. The number of lines per unit area (thethe lineline
densitydensity) is proportional to the magnitude of the electric
field.
• Therefore, the total number of lines penetrating the surface
is proportional to the product EA. This product is called the
electric flux:
ΦΦEE is electric fluxis electric flux that has units of newton-newton-
meters squared per coulomb (meters squared per coulomb (N.mN.m22
/C/C)).
ΦE is proportional to the number of electric
field lines penetrating some surface.

Electric Flux
• If the surface is not perpendicular to the
field, the flux through it must be less.
Consider figure, where the normal to the
surface of area A is at an angle θ to the
uniform electric field.
• Electric fluxElectric flux:
We can also interpret the angle as that between the electric field
vector and the normal to the surface, the product E cos θ.
Electric fluxElectric flux:
where En is the component of the electric field normal to theelectric field normal to the
surfacesurface.

Electric Flux
• Consider a general surface divided into a large number of
small elements, each of area ∆Ai. The electric flux ΦE,i
through this element is:
• The total flux through the surface:
If the area of each element approachesapproaches
zerozero, the number of elements approachesapproaches
infinityinfinity and the sum is replaced by an
integral. The electric flux is:

Electric Flux
• Consider the closed surfaceclosed surface in
figure. The vectors ∆Ai point in
different directions for the various
surface elements, but for each
element they are normal to the
surface and point outward.
• The net flux through the surface is
proportional to the net number of
lines leaving the surface.

Quiz
• Suppose a point charge is located at the center of a spherical surface.
The electric field at the surface of the sphere and the total flux through
the sphere are determined. Now the radius of the sphere is halvedhalved.
What happens to the flux through the sphere and the magnitude of the
electric field at the surface of the sphere? (a) The flux and field both
increase. (b) The flux and field both decrease. (c) The flux increases,
and the field decreases. (d) The flux decreases, and the field increases.
(e) The flux remains the same, and the field increases. (f) The flux
decreases, and the field remains the same.
If radiusIf radius rr is decreaseis decrease
thenthen EE increases.increases.
The radiusThe radius rr can becan be
eliminated from equation, iteliminated from equation, it
does not affecting the flux.does not affecting the flux.

Example: Flux Through a Cube
• Consider a uniform electric field E oriented in the x
direction in empty space. A cube of edge length l, is
placed in the field, oriented as shown in Figure 24.5. Find
the net electric flux through the surface of the cube.

Solution
• The flux through four of the faces ((3), (4), and the
unnumbered faces) is zero because E is parallel to the
four faces and therefore perpendicular to dA on
these faces.
• The net flux through faces (1) and (2):The net flux through faces (1) and (2):
• For face (1), E is constant and directed inward but dA1
is directed outward (θ = 180o
). The flux through this
face:
• For face (2), E is constant and outward and in the
same direction as dA2 (θ = 0o
). The flux through this
face:
• The net flux:The net flux:

Gauss’s Law
• Gauss’s law is a general relationship between the net electricnet electric
flux through a closed surfaceflux through a closed surface (often called a gaussian surface)
and the charge enclosed by the surfaceand the charge enclosed by the surface.
• Consider a positive charge q located at the center of a sphere
of radius r.
• The net flux through the gaussian surface is
where:where:
The flux is independent of the radius r. It depends only on the charge q
enclosed by the sphere.

Gauss’s Law
• Consider several closed surfaces surrounding a charge q as
shown in Figure 24.7.
• Surface S1 is spherical, but surfaces S2 and S3 are not.
• The number of lines through S1 = the number of lines
through the nonspherical surfaces S2 and S3.
• Therefore:
The net flux through any closed surfaceThe net flux through any closed surface
surrounding a point chargesurrounding a point charge qq is given byis given by
qq//εε00 and is independent of the shape ofand is independent of the shape of
that surface.that surface.

Gauss’s Law
• Consider a point charge located
outside a closed surface of arbitrary
shape.
• The number of electric field lines
entering the surface equals the
number leaving the surface.
• Therefore, the net electric fluxthe net electric flux
through a closed surface thatthrough a closed surface that
surrounds no charge is zerosurrounds no charge is zero.

Gauss’s Law
• Let’s see two generalized cases: (1) that of many point
charges and (2) that of a continuous distribution of charge.
• The electric field due to many charges
is the vector sum of the electric fields
produced by the individual charges.
• The electric flux through any closed
surface can be expressed as:

Gauss’s Law
• The mathematical form of Gauss’s law states
that the net flux through any closed surface is:
where E is the electric field and qin is
the net charge inside the surface.
The surface S surrounds only one charge, q1; hence,
the net flux through S is q1/ε0. The flux through S due
to charges q2, q3, and q4 outside it is zero because
each electric field line from these charges that enters
S at one point leaves it at another.

Quiz
• If the net flux through a gaussian surface is zero, the
following four statements could be true. Which of the
statements must be true? (a) There are no charges inside
the surface. (b) The net charge inside the surface is zero.
(c) The electric field is zero everywhere on the surface.
(d) The number of electric field lines entering the surface
equals the number leaving the surface.
Answer: (b) & (d)

Example
• A spherical gaussian surface surrounds a point charge q.
Describe what happens to the total flux through the
surface if (A) the charge is tripled, (B) the radius of the
sphere is doubled, (C) the surface is changed to a cube,
and (D) the charge is moved to another location inside
the surface.

Solution
the surface.
Answer
• (A) The flux through the surface is tripledtripled because flux isflux is
proportional to the amount of charge inside the surfaceproportional to the amount of charge inside the surface.
If q2 = 3q,
ΦE2 = 3ΦE

Solution
the surface.
Answer
(B) The flux does not changeThe flux does not change
because all electric field lines from the
charge pass through the sphere,
regardless of its radius.
The same number of field lines & the sameThe same number of field lines & the same
flux pass thru both of these area elementsflux pass thru both of these area elements

Solution
the surface.
Answer
(C) The flux does not changeThe flux does not change when the
shape of the gaussian surface changes
because all electric field lines from the
charge pass through the surface,
regardless of its shape.

Solution
the surface.
Answer
(D) The flux does not changeThe flux does not change because
Gauss’s law refers to the total chargetotal charge
enclosedenclosed, regardless of where the
charge is located inside the surface.

Application of Gauss’s law to
various charge distributions
• Gauss’s law is useful for determining electric fields when the
charge distribution is highly symmetric.
• The following examples demonstrate ways of choosing the
gaussian surface.
• To determine which portion of the surface satisfies one or
more of the following conditionsconditions:
1. The value of the electric field can be argued by symmetry to
be constant over the portion of the surface.
2. The dot product in Equation 24.6 can be expressed as a
simple algebraic product E dA because E and dA are parallel.
3. The dot product in Equation 24.6 is zero because E and dA are
perpendicular.
4. The electric field is zero over the portion of the surface.

Example: A Spherically Symmetric Charge
Distribution
• An insulating solid sphere of radius a has a uniform volume charge
density ρ and carries a total positive charge Q (Fig. 24.10).
• (A) Calculate the magnitude of the electric field at a point outside the
sphere.
• (B) Find the magnitude of the
electric field at a point inside the
sphere.
• What If? Suppose the radial position
r = a is approached from inside the
sphere and from outside. Do we
obtain the same value of the electric
field from both directions?
For point outside
the sphere For point inside
the sphere

Solution (A) -
Calculate the magnitude of the electric field at a point outside theCalculate the magnitude of the electric field at a point outside the
spheresphere
• To reflect the spherical symmetry, let’s choose a spherical
gaussian surface of radius r, concentric with the sphere, as
shown in Figure 24.10a. For this choice, condition (2) is
satisfied everywhere on the surface and E . dA = E dA.
• By symmetry, E has the same value everywhere on the
surface, which satisfies condition (1), so we can remove E
from the integral:
• Solve for E:

Solution (B) –
Find the magnitude of the electric field at a point inside the sphereFind the magnitude of the electric field at a point inside the sphere
• Let’s choose a spherical gaussian surface having radius r < a. Let
V’ is the volume of smaller sphere. The charge q in within the
gaussian surface of volume V’ is less than Q.
• Calculate qin:
• Apply Gauss’s law in the region r < a:

What If? –
Suppose the radial position r = a is approached from inside and
outside.
Answer
•Equation (1) shows that the electric field approaches a value
from the outside given by
•From the inside, Equation (2) gives
Therefore, the value of the field is the same asthe value of the field is the same as
the surface is approached from both directionsthe surface is approached from both directions.
A plot of E versus r is shown in Figure 24.11.
Notice that the magnitude of the field is
continuous.

Example: A Cylindrically Symmetric Charge
Distribution
• Find the electric field a distance r from a line of positive
charge of infinite length and constant charge per unit
length λ (Fig. 24.12a).

• Let’s choose a cylindrical gaussian surface of radius r and
length l, that is coaxial with the line charge.
• The flux through the ends of the gaussian cylinder is zero
because E is parallel to these surfaces.
• Noting that the total charge inside gaussian surface: qin = λl,
then
Solution

What If?
• What if the line segment in this example were not infinitely long?
Answer
• If the line charge in this example were of finite length, the electric field
would not be given by Equation 24.7not be given by Equation 24.7. A finite line charge does not
possess sufficient symmetry to make use of Gauss’s law because the
magnitude of the electric field is no longer constant over the surface of
the gaussian cylinder: the field near the ends of the line would be
different from that far from the ends. Therefore, condition (1) would
not be satisfied in this situation.
• Furthermore, E is not perpendicular to the cylindrical surface at all
points: the field vectors near the ends would have a component
parallel to the line. Therefore, condition (2) would not be satisfied.
• For points close to a finite line charge and far from the ends, Equation
24.7 gives a good approximation of the value of the field.

Example: A Plane of Charge
• Find the electric field due to an infinite plane of positive
charge with uniform surface charge density σ.

Solution
• Notice that the plane of charge is infinitely large. Therefore, the electricthe electric
field should be the same at all points equidistant from the planefield should be the same at all points equidistant from the plane.
• The flux through each end of the cylinder is EA; hence, the total flux
through the entire gaussian surface is just that through the ends,
ΦE = 2EA
Noting that the enclosed charge is qin = σA:
• Solve for E:
We conclude that E = σ/2ε0 at any distance from the plane.
The field is uniform everywhere.
Figure 24.14 shows this uniform field due to an infinite
plane of charge, seen edge-on.

What If?
• Suppose two infinite planes of charge are parallel to each other, one
positively charged and the other negatively charged. The surface charge
densities of both planes are of the same magnitude. What does the
electric field look like in this situation?
Answer
It first addressed in the What If? section of
Example 23.9 (but not in the syllabus).
The electric fields due to the two planes add in
the region between the planes, resulting in a
uniform field of magnitude, E = σ/ε0.
Figure 24.15 shows the field lines for such a
configuration. This method is a practical way to
achieve uniform electric fields with finite-sized
planes placed close to each other.

Electric PotentialElectric Potential and PotentialPotential
DifferenceDifference
• When a charge q is placed in an electric field E, there is an
electric force: Fe = qE acting on the charge.
• For an infinitesimal displacement ds of a point charge q
immersed in an electric field, the work done is
Wint = Fe . ds = qE . ds.
• Internal work done in a system is equal to the negative of
the change in the potential energy of the system:
Wint = -∆U.
• Therefore, the electric potential energy of the charge–field
system is changed by an amount:

• For a finite displacement of the charge from some point (A) in space
to point (B), the change in electric potential energychange in electric potential energy of the system is
• Dividing the potential energy by the charge gives the electric
potential (or simply the potential) V:
• The potential difference: ∆V = VB - VA between two points A and B in
is the change in electric potential energy (change in electric potential energy (∆U) of the system when
a charge q is moved between the points divided by the charge (divided by the charge (qq)):

• If the agent moves the charge from A to B without changingwithout changing
the kinetic energythe kinetic energy, the agent performs work that changes the
potential energy of the system: W = ∆K + ∆U W = ∆U.
• The work done by an external agent in moving a charge q
through an electric field at constant velocity is:
• SI unit of both electric potential and potential difference is
Joules per CoulombJoules per Coulomb, defined as a volt (V). 1 V = 1 J/C
• SI unit of electric fieldelectric field (N/C) can also be expressed in volts per
meter: 1 N/C = 1 V/m
• A unit of energy commonly used in atomic and nuclear physics
is the electron volt (eV),

Quiz
• In Figure 25.1, two points A and B are located within a region
in which there is an electric field. (i) How would you describe(i) How would you describe
the potential differencethe potential difference ∆∆VV == VVBB -- VVAA?? (a) It is positive. (b) It is
negative. (c) It is zero. (ii) A negative charge is placed at A andii) A negative charge is placed at A and
then moved to Bthen moved to B. How would you describe the change in
potential energy of the charge–field system for this process?
Choose from the same possibilities.
Answer
• (i) (b)
• (ii) (a)
Whether the test charge is positive (+) or negative
(-), the following general rules apply: UU increasesincreases if
the test charge moves in the direction opposite the
electric force; UU decreasesdecreases if moves in the same.
VB < VA
∆V = (-)

Potential Difference in a Uniform
Electric Field
• Let’s calculate the potential difference between two points A and B separated
by a distance d, where the displacement s points from A toward B and is parallel
to the field lines.
• Because E is constant, it can be
removed from the integral sign,
which gives
The negative sign indicates that theThe negative sign indicates that the
electric potential at point B is lowerelectric potential at point B is lower
than at point A; that is,than at point A; that is, VVBB << VVAA..

Electric Field
• Now suppose a charge q moves from A to B, the change inchange in
the potential energythe potential energy of the charge–field system
• IfIf qq is positive, thenis positive, then ∆∆UU is negativeis negative.
• In a system consisting of a (+) charge and an electric field, the
electric potential energy of the system decreases when the charge
moves in the direction of the field.
• If a (+) charge is released from rest in this electric field, it
experiences an electric force in the direction of E.
• IfIf qq is negative, thenis negative, then ∆∆UU is positive and the situation is reversed.is positive and the situation is reversed.
• A system consisting of a (-) charge and an electric field gains electric
potential energy when the charge moves in the direction of the
field.
• If a (-) charge is released from rest in an electric field, it accelerates
in a direction opposite the direction of the field.

Electric Field
• Now consider a charged particle that moves between A and B in a
uniform electric field such that the vector s is not parallel to the field
lines. Then,
The change in potential energy of the charge–field system is:
The potential difference VB - VA is equal to the
potential difference VC - VA. Therefore, VB = VC.
The name equipotential surface is given to any
surface consisting of a continuous distribution of
points having the same electric potential.
The equipotential surfacesequipotential surfaces associated with a
uniform electric field consist of a family of parallela family of parallel
planes that are all perpendicular to the fieldplanes that are all perpendicular to the field.

Example: The Electric Field Between Two Parallel
Plates of Opposite Charge
• A battery has a specified potential difference ∆V between its terminals
and establishes that potential difference between conductors attached
to the terminals. A 12-V battery is connected between two parallel
plates as shown in Figure 25.5. The separation between the plates is d
= 0.30 cm, and we assume the electric field between the plates to be
uniform. (This assumption is reasonable if the plate separation is small
relative to the plate dimensions and we do not consider locations near
the plate edges.) Find the magnitudemagnitude of the electric field between the
plates.

Solution
• Use Equation 25.6 to evaluate the magnitude of the electric field
between the plates:
• The configuration of plates in Figure 25.5 is called a
parallel-plate capacitor.
+-
E

Example: Motion of a Proton in a Uniform
Electric Field
• A proton is released from rest at point A in a uniform electric
field that has a magnitude of 8.0 x 104
V/m (Fig. 25.6). The
proton undergoes a displacement of magnitude d = 0.50 m to
point B in the direction of E. Find the speed of the proton after
completing the displacement.

Solution
• The conservation of energy equation, for the isolated
system of the charge and the electric field:
• Substitute the changes in energy for both terms:
• Solve for the final speed of the proton and substitute for ∆V from
Equation 25.6:

Electric Potential and Potential Energy
Due to Point Charges
• The general expression for potential difference:
• Therefore
• By substitutions, we find:
• Hence, the potential difference:

Electric Potential and Potential
Energy Due to Point Charges
• It is customary to choose the reference of
electric potential for a point charge to be
VVAA = 0 at= 0 at rrAA == ∞∞.
• Therefore, the electric potential (V) due to
a point charge at any distance (r) from the
charge (q) is

• The total electric potential at some point P due to several point
charges is
• If an external agent brings a chargea charge qq22 from infinity to pointfrom infinity to point PP, the
work done is
W = q2∆V
The electric potential energyelectric potential energy of aa
pair of point chargespair of point charges::

• If the system consists of more than two charged particles,
we can obtain the total potential energy of the system by
calculating U for every pair of charges.
• For example, the total potential energy of the system of
three charges:three charges:

Quiz
Answer
• (i) : (c)
• (ii) : (a)
If the charges are of the same sign, then U is
positive. Positive work must be done by an external
agent on the system to bring the two charges near
each other (because charges of the same sign
repel).

Example
The Electric Potential Due to Two Point Charges
• As shown in Figure 25.10a, a charge q1 = 2.00 µC is located at
the origin and a charge q2 = -6.00 µC is located at (0, 3.00) m.
• (A) Find the total electric potential due to these charges at the
point P, whose coordinates are (4.00, 0) m.
• (B) Find the change in potential energy of the system of two
charges plus a third charge q3 = 3.00 µC as the latter charge
moves from infinity to point P (Fig. 25.10b).

The capacitance C of a capacitor is :
Definition of Capacitance
• Consider two conductors is called a capacitor. The conductors
are called plates. If the conductors carry charges of equal
magnitude and opposite signopposite sign, a potential difference ∆V exists
between them.
Charge Q is:
This relationship can be written as
The SI unit of capacitance is the farad (F): 1 F = 1 C/V

Calculating Capacitance
• Although the most common situation is that of two
conductors, a single conductor also has a
capacitance.
• For a single sphericalsingle spherical charged conductor, the
potential of the sphere of radius a is V = keQ/a, and
setting V = 0 for the infinitely large shell (infinite
radius), then

Parallel-Plate Capacitors
• Two parallel, metallic plates of equal area A
are separated by a distance d as shown in
Figure.
• One plate carries a charge +Q , and the
other carries a charge -Q. The surfacesurface
charge densitycharge density on each plate is σ = Q/A.
• The capacitance is

Quiz
Answer: (d)
• Charge Q is linearly proportional to the potential difference.

Quiz
Answer: (a)
When key is pushed down then d
decreases. So, if
dd decreases thendecreases then CC is increasesis increases..

Electrostatics - 203PHYS

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