Sindh Mdcat Exclusive Course Physics Electrostatics — Solved MCQs

Question 1

Electric field strength at a point between oppositely charged plates is 'E'. If the distance between plates is reduced to half, what will be the new value of electric intensity?

Accepted Answer

D. 2E — Between two parallel plates, the electric field strength is given by:E = V/dIf the distance is halved, electric field strength is doubled since both are inversely related.

Question 2

When a charged particle is projected perpendicular to a magnetic field:

Accepted Answer

A. Its path is circular in a plane perpendicular to the plane of magnetic field — When a charged particle is projected perpendicular to a magnetic field,The path of particle is circular in a plane perpendicular to the plane of magnetic field.The correct option is A.

Question 3

A parallel plate capacitor has a capacitance C. If the distance between the plates halves, the capacitance will be?

Accepted Answer

D. 2 C — The capacitance is inversely proportional to distance between the plates. Halving the distance would double the capacitance giving us 2C.

Question 4

The energy stored in the capacitor is written as:

Accepted Answer

A. ½ CV2 — The energy stored in a capacitor is given by the formula:U = 1/2 CV2where U is the energy stored in the capacitor, C is the capacitance of the capacitor, and V is the potential difference across the capacitor.

Question 5

The energy supplied by the cell to the charge carriers is derived from the conversion of:

Accepted Answer

B. Chemical energy into electrical energy — The correct answer is that the energy supplied by the cell to the charge carriers is derived from the conversion of chemical energy into electrical energy. This process occurs in both primary and secondary cells, where chemical reactions within the cell generate electrical energy that can be used to power devices. Option A is incorrect because endothermic reactions and the conversion of heat energy into chemical energy are not the processes occurring in conventional electrical cells. Option C is incorrect as it describes the function of solar cells, which are distinct from the electrochemical cells typically used in circuits. Option D is incorrect because converting mechanical energy into electrical energy is characteristic of generators, not cells or batteries used in standard electrostatic applications.

Question 6

Which statement describes the electric potential difference between two points in the electric field of charge Q?

Accepted Answer

C. The work done in moving a test charge between points divided by the magnitude of the test charge. — This question is simply revising how the electric potential is defined as it is work done per unit charge in moving that charge from one place to the other in the electric field.All other options are not the definitions of electric potential.

Question 7

Two capacitors C1 = 2µ and C2 = 4µ F are connected in series across in a 100V supply. Find the effective capacitance.

Accepted Answer

D. 4/3 µF — Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)Therefore,1/Ceffective = 1/C1 + 1/C21/Ceffective = ½ +¼ 1/Ceffective = ¾C effective= 4/3 µF Hence option D is the correct answer.

Question 8

What is true about the electric field and electric force?

Accepted Answer

C. Electric field lines are towards the negative and electrons flow in the opposite direction. — Electric field lines always point away from a positive charge and towards a negative point. In fact, electric fields originate at a positive charge and terminate at a negative charge.This shows that OPTION B and D ARE INCORRECT. Electrons flow from lower potential terminal to higher potential terminal being negatively charged particles it flows from negative to positive poles hence it is the opposite direction as the electric field.So,OPTION C is incorrect.

Question 9

An electron is situated midway between two parallel plates 0.5 cm apart. One of the plates is maintained at a potential of 60 volts above the other. The force on the electron is: (e = -1.6x10-19C)

Accepted Answer

C. 1.92x10-15 N — To solve this problem, we first calculate the electric field (E) between the parallel plates using the formula E = V/d, where V is the potential difference and d is the distance between the plates. Here, V = 60V and d = 0.5 cm = 0.005 m. Thus, E = 60V/0.005m = 12,000 V/m.Next, we apply the formula for the force on a charge in an electric field: F = eE. The charge of an electron is e = -1.6x10-19 C. Therefore, F = (1.6x10-19 C)(12,000 V/m) = 1.92x10-15 N.Option A is incorrect due to a possible miscalculation of E. Option B underestimates the force, likely from an error in applying the formula. Option D overestimates the force, possibly by using incorrect values. Option E significantly overestimates the force, indicating a misunderstanding of the relationship between V, d, and the resulting force.

Question 10

By increasing the area of the plates and decreasing the distance between them, the capacitance of a capacitor:

Accepted Answer

A. Increases — Capacitance is directly proportional to the electrostatic force field between the plates. This field is stronger when the plates are closer together. Therefore, as the distance between the plates decreases, capacitance increases. Moreover, the same applies when we increase the surface area of the plates. Increasing the surface area of the plates would incite an increase in capacitance.

Question 11

A sphere of charge +Q is fixed in a position. A smaller sphere of +q is placed near the larger sphere and released from the rest. The small sphere will move away from the large sphere with?

Accepted Answer

D. Increasing velocity and decreasing acceleration — In this scenario, the small sphere (+q) is initially at rest near a larger fixed sphere (+Q). Because both spheres are positively charged, they repel each other. Initially, the repulsive force causes the small sphere to accelerate, increasing its velocity. However, as the sphere moves further away, the force of repulsion decreases in magnitude due to Coulomb's Law (F ∝ 1/r2), which causes the acceleration to decrease. Therefore, the correct answer is that the small sphere moves away with increasing velocity and decreasing acceleration.The other options are incorrect because:Decreasing velocity and decreasing acceleration: The velocity increases due to initial acceleration.Decreasing velocity and increasing acceleration: Acceleration decreases as the distance increases.Decreasing velocity and constant acceleration: Neither velocity decreases nor acceleration remains constant.Increasing velocity and increasing acceleration: Acceleration does not increase; it decreases as the force diminishes.

Question 12

Two capacitors C1=3 uF and C2=6 uF are in series across a 90 volts D.C. supply. The total capacitance is given by:

Accepted Answer

B. 2 uF — 1/C1 + 1/C2 = 1/C⅓ + ⅙ = 1/C2 + 1 /6 = 1/C3/6 = 1/C½ =1 /CC = 2 uF

Question 13

If the potential at a point which is 1m from a charge is 1 volt, then the potential at a point which is 2m from the same charge will be:

Accepted Answer

C. 0.5 V — Electric potential is given by V= kQ/rAs V is inversely proportional to r so increasing the R by a factor of 2 (1m to 2m), the electric potential, subsequently, drops by a factor of 2 (1 V -> 0.5 V).

Question 14

If a dielectric material is placed between two plates of a capacitor, the net capacitance of the capacitor:

Accepted Answer

B. Increases — ○The strength of the electric field is reduced due to the presence of dielectric. If the total charge on the plates is kept constant, then the potential difference is reduced across the capacitor plates. In this way, dielectric increases the capacitance of the capacitor.So, option A,C,D is wrong .

Question 15

The values of electric intensity will _ due to the presence of dielectric medium:

Accepted Answer

C. Decrease — The strength of the electric field is reduced due to the presence of dielectric. If the total charge on the plates is kept constant, then the potential difference is reduced across the capacitor plates.

Question 16

Ohm multiplied by farad is equivalent to:

Accepted Answer

A. Time — The formula for charging and discharging capacitor is RC=TR: resistance C: capacitanceT: time for charging/ discharging

Question 17

“The magnitude of electrostatic force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.” This is known as:

Accepted Answer

E. Coulomb's Law — The statement given is a direct description of Coulomb's Law. This law quantitatively defines the electrostatic force between two point charges as F = k(q1*q2)/r², where F is the force, q1 and q2 are the magnitudes of the charges, r is the distance between the charges, and k is Coulomb's constant. This law is fundamental in understanding electrostatic interactions.Other options explain different physical laws: Gauss's Law pertains to electric fields, Ohm's Law to electrical circuits, Faraday's Law to electromagnetic induction, and Newton's Laws to mechanics. None of these describe the electrostatic force between point charges.

Question 18

The net charge flowing through any section of the conductor per unit time is known as?

Accepted Answer

C. Electric current — The correct answer is Electric current. Electric current is the rate at which electric charge flows past a point or through a region. It is measured in amperes (A), where 1 ampere equals 1 coulomb of charge passing through a point in one second.Other options explained: Electric intensity pertains to the strength of an electric field, not to the flow of charge. Electric flux deals with the field lines through an area, not the charge flow through a conductor. Electric potential is the work needed to move a charge in an electric field and does not measure flow. Dielectric constant relates to a material's ability to transmit electric fields, not to charge flow.

Question 19

If the potential difference of 1 volt is applied across the end of a conductor and the current flowing through the conductor is one ampere, then the resistance of the conductor is said to be?

Accepted Answer

C. 1 Ohm — According to Ohm's Law, V = IR, where V is the potential difference, I is the current, and R is the resistance. Given a potential difference (V) of 1 volt and a current (I) of 1 ampere, the resistance (R) is calculated as R = V/I = 1 volt / 1 ampere = 1 ohm. Therefore, the correct answer is 1 Ohm.Options involving farads and coulombs are incorrect because they are units for capacitance and electric charge, respectively, not resistance. The option of 10 ohms is incorrect as it does not match the calculated resistance using the given values.

Question 20

A particle carrying charge of 2e falls through a potential difference of 3.0 V. Calculate energy acquired by it.

Accepted Answer

A. 9.6 × 10-19 J — To find the energy acquired by the particle, use the formula: Energy = qV, where q is the charge and V is the potential difference. The charge of the particle is 2e, where e is the elementary charge, approximately 1.6 × 10-19 C. The potential difference is given as 3.0 V. Thus, Energy = 2e × 3 V = 6eV. Converting eV to joules gives: Energy = 6 × 1.6 × 10-19 J = 9.6 × 10-19 J.Options B, C, D, and E are incorrect because they do not follow the correct calculation using the formula and given values.

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