Generate Quiz 6A2CD945545E2

This question is a high-yield concept because it has been asked repeatedly in other papers.

With the Greek words xeros meaning “dry” and graphos meaning “writing,” the word “xerography” refers to an electrostatic process used in the production of copy machines.

By measuring the force applied to a unit positive charge positioned at a given location, one can determine the electric field intensity. Consequently, E = F/q is the formula.

Using the following formula, we can easily determine how many electrons must be removed from a body in order to produce one coulomb of charge:
‘Q’ stands for the charge, ‘n’ for the number of electrons, and ‘e’ for the electronic charge in the formula Q = ne (*).

1.6 * 10⁻¹⁹ C = 1 Coulomb e = Charge ‘Q’
Then, using (*), we get 1 = n * 1.6 * 10⁻¹⁹ 1/1.6 * 10⁻¹⁹ = n 6.25 * 10¹⁸ = number oof electrons
Therefore, 6.25 * 10¹⁸ electrons must be removed from a body in order to generate one coulomb of charge.

The formula for electric potential is V=kQ/r.

Since V and r are inversely proportional, increasing R by a factor of two (1 m to 2 m) causes the electric potential to decrease by a factor of two (1 V-> 0.5 V).

Capacitors are designed to store charge.

The SI unit for free space permittivity is C2N−1m−2.

F/q = E

F = 2 N, q = 3 C

Electromagnetic field = 2N / 3C, then

Each capacitor in a parallel configuration receives the entire battery voltage. As a result, the potential difference across each capacitor is equal.

The joule is the unit of energy, the Newton per Coulomb is the unit of electric field strength, and the Tesla is the unit of magnetic field strength. Joules per Coulomb is the unit of electric potential difference.

Coulomb’s law states that a force is repulsive (positive F) when the signs of the charges are the same and attractive (negative F) when they are opposite.

F=k q1q2/r2

The charge distribution and system geometry determine the electric intensity at the center of a uniformly distributed charge. The electric field intensity is zero at the center of a spherically symmetric charge distribution, like a uniformly charged sphere. This is because the symmetry of the distribution causes the electric field vectors from every point on the sphere to cancel each other out.

Electrostatics is the study of how charges behave when they are at rest.

A or B are the only options with the same charge throughout the circuit, so either of these answers is correct.

Q = CV

Series capacitance can be computed as follows:

1/C = 1/C1 + 1/C2

1/C = 1/12 + 1/24

1/C = 2+1/24

1/C = ⅛

C is equal to eight microfarads.

->Q=CV

Q = (8×10^-6)(360)

Q = 2880 x 10-6 C

The electric field strength between two parallel plates is determined by:

E = V/d

Due to their inverse relationship, electric field strength doubles when distance is cut in half.

According to option D, depending on the motion of the charge and the presence of other charges or magnetic materials, all three of the aforementioned fields—static, electric, and magnetic—can develop when a charge encounters a force. A charge creates an electric field (static field) when it is at rest. It produces both magnetic and electric fields when moving. Thus, this choice is appropriate.

Option C is accurate since increasing the area of the plates will allow it to hold more charge in order to double the charge.

Capacitance is therefore not determined by charge. The amount of charge that a capacitor can store is determined by its capacitance.

Giving a double charge won’t double the capacitance. This additional fee will be squandered.

One electron, which is equivalent to the charge of one proton, is the smallest indivisible unit of charge. Thus, choice C.
An electron has a minimum charge of 1.6 x 10-19 C.

5.3 x 10^-5 C = Q

V equals 6.0 V.

Applying Q = CV, we obtain:

C = Q/V = (5.3 x 10^-5 C)/(6.0 V) = 8.83 x 10^-6F

Let’s now determine how much charge the capacitor stores when it is connected to a 9.0 V battery. We possess:

V equals 9.0 V.

C = 8.83 x 10^-6 F

Applying Q = CV, we obtain:

Q = CV = (8.83 x 10^-6 F) x (9.0 V) = 7.95 x 10^-5 C

A capacitor’s capacitance indicates how much electric charge it can hold. The magnitude of the electric charge stored on one of a capacitor’s plates divided by the potential difference (voltage) across the plates is known as the capacitance. It is measured in farads (F) and is represented by the letter C.

V/d = E

Note that the distance should be measured in meters. Thus, 2 cm = 0.02 m.

20-(-20)/0.02

= 2000 V/m

First, F=4πϵ0​1​r2q1​q2 is the force between charges, say q1 and q2, separated with distance r.Step 2: Calculate the new force

When both the charge and the distance between charges double, for example, q1′ = 2q1, q2′ = 2q2, and r′ = 2r
This will be the new force:
F′=4πϵ0​1​r′2q1′​q2′​​F′=4πϵ0​1​(2r)2(2q1​)(2q2​)​=4πϵ0​1​r2q1q2 = F
As a result, the force will not change.

Current is the net charge flow through a conductor in a unit of time. Amperes are used to measure current.

When the electric field is uniform, the two electric field lines are

1. straight

2. in parallel

3. equally spaced

A field where the field strength stays constant is called a uniform electric field.

The work required to move a unit positive charge from infinity to a given location is known as the electric potential at that location. The amount of “work done in moving a unit positive charge from one point to the other against the field” is then used to define the potential difference between two points.

Volts per meter (V/m) is the unit of measurement for electric field strengths (E).

According to Coulomb’s Law, the electrical force between two charged objects is inversely proportional to the square of their distance from one another and directly proportional to the product of the charges on the objects. Therefore, Option D is the right response to this query.

Force = k×q×q/r²

An electrostatic field between the capacitor’s plates stores energy. The work that the battery does is one way to express this energy.

One coulomb of charge contains 6.25×1018 electrons.

A parallel capacitor can store 12 qv of energy, where q = cv, or 1/2 cv2.

Electric field lines are always directed toward a negative point and away from a positive charge. A positive charge is where electric fields begin, and a negative charge is where they end. This demonstrates that choices B and D are not appropriate.

Since electrons are negatively charged particles that move from the lower potential terminal to the higher potential terminal, they flow in the opposite direction of the electric field—from the negative to the positive poles. Option C is therefore accurate.

During that era, Charles Carlson recognized the need for a simple and rapid method of copying. In 1937, he invented electrophotography, which was later renamed Xerography, derived from the Greek words xeros, which means dry, and graphos, which means writing.

Correct response.

Given:

d=0.5cm=0.005m

V = 0.6V

e = -1.6×10-19C

F=?

Answer:

E=v/d, as

60/0.005

= 12,000 V/m

E=12,000V/m, then

E=F/e now

F = eE

F = (1.6×10-19C)(12,000V/m)

F = 1.92×10-15N

Outcome:

F = 1.92×10-15N

An object’s electric charge—whether positive or negative—determines how it will react to an electric or magnetic field. Coulombs, or Cs, are the SI unit of charge.

A scalar quantity is electric potential. The electric potentials of the two positively charged point charges will be added at their midpoint, doubling the potential.

The answer

One meter, one meter

V=?———-+q——–V=?

V = k q/r + k q/r

V = k q/r + k q/r

V = 2 k q/r.

The SI unit of capacitance is a farad. It is the quantity of electric charge that a capacitor can hold per unit of voltage. For every volt of potential difference, one farad is equivalent to one coulomb of electric charge. Coulomb/volt, then.

The Coulomb force F applied per unit positive electric charge q at a given location can be used to determine the strength of an electric field E at that location; this can be expressed simply as E = F/q.

In general, energy density is expressed as ½ εo E2.

Nevertheless, the permittivity of free space εo will be multiplied by the relative permittivity of the medium εr if the medium differs from free space.

As a result, the capacitor’s energy density can now be expressed as ½ εoεr E2. Option B is therefore accurate.

The electric field of a capacitor has the capacity to store energy.

An electric field forms between the plates of a capacitor when it is charged, accumulating electric potential energy. Because the positive and negative charges on the capacitor plates are separated, the energy is stored in the electric field. When the capacitor discharges, the energy is released, and charges move from one plate to the other, creating an electric current flow.

Projecting a charged particle perpendicular to a magnetic field causes

In a plane perpendicular to the magnetic field plane, the particle’s path is circular.

A is the right choice.

Gaussian surface: In three dimensions, the Gaussian surface is a closed surface that is used to compute a vector field’s flux. The gravitational field, electric field, or magnetic field are examples of these vector fields.

The electric field inside the capacitor is reduced when a dielectric is present between the two plates.

Because C is inversely proportional to distance, capacitance will increase three times for every three decreases in distance.

The capacitance rises and the plate separation decreases when the key is pressed. Capacitance is independent of the external circuit and solely depends on the capacitor’s construction.

When an electron is released from an electric field, it accelerates in the opposite direction of the magnetic field.

Coulomb’s Law is defined as follows: the electrostatic force (F) between two point charges is inversely proportional to the square of their distance (r2) and directly proportional to the product of their charge magnitudes (qq). In terms of mathematics:

F = kqq/r2

where k is the proportionality constant.

Option E is therefore the right one.

Here, the precise charge is being requested. The following is the formula for that:

Q/m {where m is the mass and Q is the electron’s charge}

Enter m=9.1×10-31 and Q=1.6×10-19 into the formula:

=1.6×10-19/9.1×10-31

1.76×1011

The dielectric that is positioned between the capacitor’s plates weakens the electric field between them, which causes a slight voltage to exist between them for the same charge. For a lower voltage, the capacitor can hold more charge. As a result, the capacitance rises.

Since they are electrically neutral, neutrons are chargeless. This is because their protons and electrons, which have opposing charges, are equal in number. An electron has a charge of -1, while a proton has a charge of +1. Thus, a neutron’s charges cancel each other out when added together, leaving a net charge of zero.

A is the right choice.

E = ½ Q V is the amount of energy stored in the capacitor.

= ½ × 27 × 10-6 × 18

= 2.43×10-4 J

There can only be one response because this is a numerical question.

“Electrostatic induction”—also referred to as “electrostatic influence” or just “influence”—is the redistribution of electric charge within an object brought on by the influence of nearby charges.

Positive charges are repelled by one another, so the electric field surrounding a single positive charge is ‘radially outward’. Electric fields are shown as beginning on positive charges and ending on negative charges when they are represented by lines of force, also known as field lines.

In a uniform or non-uniform electric field, an electric dipole always experiences a torque. However, a dipole will also feel a net force of attraction in an uneven electric field. Thus, in a non-uniform electric field, the electric dipole experiences both force and torque.

In series are C1 and C2.

C1 and C2’s capacitance (C12) will be,

C12 = 1/C1 + 1/C2

C12 is equal to 1/2 + 1/2.

C12 = 1C

C3 and C1 and C2 are parallel.

Total Capacitance will therefore be

CTOTAL is equal to C12 + C3.

CTOTAL = 1 + 2

CTOTAL = 3C

RC=T is the formula used to charge and discharge capacitors.

R: resistance

C: capacitance

T: charging/discharging time

As stated on page 40 of the Physics Sindh Textbook.

The statement “….. two static point charges…..” makes it very evident that the charge is at a single location in space.

E’s formula:

E = q / 4πɛr2

Given that

E1 and E2 are equal.

q2 / 4πɛr22 = q1 / 4πɛr12

q2 / 4πɛ (2 r1)2 (r2 = 2r1) = q1 / 4πɛr12

q1 / 4πɛr12 = q2 / 16 πɛ r12

q1 / 4 = q2 / 16

q2 = 4q1

Therefore, choice A is the right one.

The following formula provides the energy stored in a capacitor:

U = 1/2 CV2

where V represents the potential difference across the capacitor, C represents the capacitance of the capacitor, and U represents the energy stored in the capacitor.

Because of the dielectric, the electric field’s strength is diminished. The potential difference across the capacitor plates is lessened if the total charge on the plates remains constant. The capacitor’s capacitance is thereby increased by the dielectric.

An object’s minimum charge is 1.6×10^-19 C.

In free space, the permittivity is roughly 8.85 x 10-12 C2 /(N.m2). Option A is correct because this is a fact.

The charge stays the same in series. Options C and D are thus nullified.

In order to determine total capacitance:

1/C = 1/12 + 1/24

C = 8F

C = Q/V

8 x 36 = 288C

Thus, option A is the right response.

The formula for a spherical capacitor’s capacitance is 4πε (ab)/(b-a).

where an is the separation between the inner surface and the center.

The distance b between the center and the outer surface

ε = 8.855 × 10-12 m-3 kg-1 s4 A2

Or ε = 8.855 × 10-12 C2 / N . m2

The formula for a spherical capacitor’s capacitance is 4πε (ab)/(b-a).

where an is the separation between the inner surface and the center.

The distance b between the center and the outer surface

ε = 8.855 × 10-12 m-3 kg-1 s4 A2

Or ε = 8.855 × 10-12 C2 / N . m2

All current sources transform non-electrical energy, including heat, mechanical, chemical, and solar energy, into electrical energy. Current comes from a wide variety of sources.

Since the field is created between the two charges—from the +ve charge to the -ve charge—no electric field can exist on X or Z, and the field lines will only cross at Y.

The particle’s direction is always relative to the lines of the electric field.

Given:

Charge(q)=2e-

3.0V is the potential difference (V).

Locate:

Energy =?

Answer:

Energy equals completed work.

Work = qV

Energy = 2e×3 = 6eV

Energy = 6×1.6×10-¹⁹

Energy=9.6×10-¹⁹J

An electric field moves in a positive-to-negative direction. Because it is drawn to the positive side, a negatively charged electron will accelerate in the opposite direction of the electric field.

According to Coulomb’s law, the force of attraction or repulsion between two electric charges acting in a straight line is inversely proportional to the square of their distance from one another and directly proportional to the product of their charges.

One name for the force per unit charge is:

F/q = E. This formula relates to the intensity of an electric field.