Generate Quiz 6A2CD93C1FC27

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

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.

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.

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.

Voltage or potential difference is the amount of work done per unit of charge.

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.

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.

A parallel plate capacitor’s capacitance is directly proportional to the dielectric constant (k) of the medium between the plates and the area of one of the plates (A). It also has an inverse relationship with the separation between the plates (d).

In terms of mathematics,

(εo is the free space permittivity, which has a value of 8.85 x 10-12 F/m).

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 electric potential difference is the difference in the electric potential energy of a unit charge between two points in the electric field.

The capacitance of a capacitor is increased when a dielectric material is placed between its plates. The material’s relative permittivity or dielectric constant (k) controls how much of this increase occurs. The capacitance of the capacitor with the dielectric material divided by the capacitance of the same capacitor without the dielectric material is known as the dielectric constant. In mathematics, C’=k×C, where C is the capacitance without the dielectric, is the capacitance with the dielectric (C’). According to this formula, the presence of a dielectric material causes the capacitance to increase by a factor of k. As a result, option A) Increases by k correctly depicts how a dielectric affects capacitance.

A is the right choice.

In line with Coulomb’s law:

F1 = k/d2.

K = F1 × d2

F2 = k /(d/2)2

F2 = 4F

There is only one right answer because this is a numerical question.

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 following formula provides the energy gained by a particle with charge q when it accelerates through a potential difference V:

U = qV

Option D is the right choice because it displays this.

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

Gauss’s law states that the electric flux through a closed surface is directly proportional to the total electric charge contained within it. It is therefore solely dependent on the enclosed charge.

E.A. = Q/ε0

Electric flux is equal to E.A.

Q is the total charge.

ε0 = dielectric constant

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

A positive charge feels a force when it is exposed to an electric field. A positive charge experiences a force that is parallel to the lines of the electric field and points in the same direction as the field. Given that positive charges are repelled from areas of lower electric potential and attracted to areas of higher electric potential, this makes sense. The positive charge feels a force in the direction of the electric field since the electric field points in the direction of decreasing potential.

The net electric field is the vector sum of the electric fields caused by +q1, -q, and q2 at any point on the spherical Gaussian surface.

Since the net charge surrounding the Gaussian surface is zero, this should not be confused with the electric flux, which is zero (net) flowing over it.

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.

The electrostatic potential varies as we move from the positive charge toward the negative one; at any given point, the electric field is negative of the potential gradient. A declining potential is indicated by a negative sign.

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.

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.

They move apart because they have the force of repulsion because they are both charged with the same thing. They do experience acceleration as a result of the force acting upon them, which raises the velocity. However, the acceleration decreases as the force acting on both objects decreases due to the growing distance between the two charges. Therefore, Option D is the right one.

Given that the electric field lines from a point charge diverge from one another, Option B is accurate. As seen in the picture below, it is a fact:

The equivalent capacitance is the total of the individual capacitances when capacitors are connected in parallel.

According to this statement, the force that object A applies to object B (FAB) is equal to the force that object B applies to object A (FBA), but it is applied in the opposite direction. The force between two charges is the same, but they act in opposite directions, according to Coulomb’s law. Option B is therefore the right choice.

The farad is the unit of capacitance. The units of voltage, current, and time are, respectively, volts, amperes, and seconds.

Electric potential is defined as energy per unit charge.

V = E/q

Therefore, choice C is the right one.

The electrostatic force’s work is solely dependent on its starting and ending positions and is not affected by the path it takes. Thus, electrostatic force is a force that is conservative.

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

q = 3e

q = 3 (1.6 x 10-19)

V is equal to 2V.

E equals qV.

E = (3 х 1.6 x 10-19)(2V)

E = 9.6 x 10-19 J

Option B is the right choice because this is a numerical question with only one possible response.

The result of connecting these three capacitors in series is one microcoulomb.

1/1 + 1/C2 + 1/C3

= ⅓ + ⅓ + ⅓

= (1+1+1) / 3.

= 3 / 3.

1/C = 1.

C = 1

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.

Keep in mind that 6.25 × 1018 electrons carry one coulomb of charge.

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.

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.

While objects B and C repel one another, indicating that they carry the same charges, objects A and B attract one another, indicating that they are oppositely charged. Objects B and C would be attracted to one another if they were both charged. As a result, the charges of objects A and C are opposite.

Since Point C has no charge, there can be no electric field there. Points A and B can never be zero because the electric field is away from the point when the charge is positive and vice versa.

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.

Because of forces of attraction and repulsion, the field lines of charged particles are never perfectly round, but they are always curved.

The distribution of charges between two objects becomes less uniform when there is frictional force between them; that is, the net charge on the objects’ surfaces changes from positive to negative, producing an electric field. Consider it frictional induction. Additionally, remember that only negative charges are transferred; charges cannot be created because they are already present in an object.

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

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.

V = W/q

V = P.E./q

Since the charges’ signs are opposite, only their direction changes, but the point charges’ magnitude stays the same.

One name for the force per unit charge is:

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

In electrostatics equations, the Coulomb constant, also known as the electric force constant or the electrostatic constant (abbreviated ke, k, or K), is a proportionality constant. It is equivalent to 8.9875517923(14)×109 (9*109)kg⋅m3⋅s−2⋅C−2 in SI units.

The following formula provides the energy stored in a capacitor:

1/2 × C × V2 = E

where V stands for voltage, C for capacitance, and E for stored energy.

Since V = 10 V and C = 2 uF in this instance,

E = 1/2 × 2 uF × (10 V)2 = 100 uJ

The following formula provides the charge on each capacitor plate:

Q = C × V

Since V = 10 V and C = 2 uF in this instance,

Q = 2 uF × 10 V = 20 uC

Consequently, D) Q = 20 uC and E = 100 uJ is the right response.

A spherical resistor’s capacitance can be found using:

Where C = (4πεab)/(b-a)

The charge on the inner sphere is denoted by Q=.

Free space permittivity is equal to ε.

a = radius of the inner sphere

b = radius of the outer sphere

Furthermore, Farad is the unit of capacitance.

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.

The following formula provides the capacitance of a parallel plate capacitor:

C = Aεo/d

where ε0 is the permittivity of free space, d is the distance between the plates, A is the area of each plate, and C is the capacitance.

Every other choice is wrong.

Gravitational pull (related to mass) and the applied electric field will exert forces on the droplet, causing qE to be directed upward and mg to be directed downward.
F = ma = qEo -mg = 0.

The equation becomes qEo = mg And , Eo = mg/q
Thus, only choice C is accurate.

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.

The stored energy is equal to Q²/2C, where Q is the charge and C is the constant capacitance.

Consequently, the square of charge and energy are directly proportional. Thus, energy increases four times when charge increases two times.

The total capacitance of capacitors connected in parallel is determined using

Ct = C1 + C2 + C3

Ct = 6μC

Among the provided, this yields the highest net capacitance.

The electric field should be directed toward Q as it shifts away from positive charges and toward negative charges. Since electric potential is a scalar quantity, directions should not be taken into account when adding it. The potential at point X will be zero since the charges are opposite and equal at the same distance from it.

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

Flux through a surface is proportional to the charge contained at its center, per Guass’ Law.

An object’s ability to exhibit electrostatic attraction or repulsion with other charged objects is known as its charge. While positive and negative charges attract one another, two positive or negative charges will repel one another.

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

1/C1 + 1/C2 = 1/C

⅓ + ⅙ = 1/C

1/C = 2 + 1/6

1/C = 3/6

½ = 1 /C

2 uF = C