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Top 100 Question on Electrostatic with Answer pdf
Here We Share Standard Multiple Choice On Electrostatics with PDF Format So dear students As we know Electrostatics is a branch of Physics that studies the effects produced in bodies as a consequence of their electric charges , or what is the same, the behavior of electric charges in equilibrium situation. This electrical charge is responsible for the electrostatic effects (attraction or repulsion) that are generated between the bodies that possess it. In this article we will discuss More than 30 MCQs on Electrostatics Class 12 and MCQs on Electrostatics for NEET and All Competitive. With the live learns teams MCQ on electrostatics Find out where you stand, your strong points and weak points and try to take corrective steps immediately with the helps of team live learns.
Check here’s the NCERT MCQ questions for physics class 12 Electrostatics with detailed answers in PDF format for free download in a single click . Class 12 MCQ Physics Questions with Answers were prepared based on the latest CBSE Board exam pattern. We here provide Electrostatics Class 12 Physics MCQ Questions with Answers to help students understand the concept very well.
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(a) Electrostatic force is a conservative force.
(b) Potential at a point is the work done per unit charge in bringing a charge from any point to infinity.
(c) Electrostatic force is non-conservative
(d) Potential is the product of charge and work.
(a) for any x for a given z
(b) for any y for a given z
(c) on the x-y plane for a given z
(d) all of these
(a) are closer in regions of large electric fields compared to regions of lower electric fields.
(b) will be more crowded near sharp edges of a conductor.
(c) will always be equally spaced.
(d) both (a) and (b) are correct.
(a) the electric field is uniform.
(b) the electric field is zero.
(c) there can be no charge inside the region.
(d) both (b) and (c) are correct.
(a) remain the same
(d) become zero
(a) The surface of the conductor is an equipoten-tial surface.
(b) The electric field just outside the surface of a conductor is perpendicular to the surface.
(c) The charge carried by a conductor is always uniformly distributed over the surface of the conductor.
(d) None of these.
(a) increases K times
(b) remains unchanged
(c) decreases K times
(d) increases K-1 times
(a) area of the plate is decreased.
(b) distance between the plates increases.
(c) area of the plate is increased.
(d) dielectric constantly decreases.
(a) capacitance will increase.
(b) energy stored will decrease.
(c) electric field will increase.
(d) voltage will decrease.
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(d) 2 CV2
(a) point Charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
13. Two identical conducting balls having positive charges q1 and q2 are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be
a. less than before
b. same as before
c. more than before
a. Q remains the same, C is doubled
b. Q is doubled, C doubled
c. C remains same, Q doubled
d. Both Q and C remain same
15. A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
d. Energy density
a. 3 x 109
b. 3 x 10-9
c. 1/3 x 10-9
d.1/3 x 109
Answer: c. 1/3 x 10-9
a. actual transfer of protons
b. actual transfer of electrons
c. actual transfer of neutrons
d. none of the above
a. transfer of electrons
b. transfer of protons
c. transfer of integral number of electrons
d. none of the above
a. 3.2 x 10-18C
b. -3.2 x 10-18C
c. -3.2 x 10-17C
d. 3.2 x 10-17C
Electrostatics Multiple Choice Questions With Answers Free Pdf
b. cylindrical symmetric
c. spherical symmetric
d. none of the above
Answer :- asymmetric in nature.
a. spherical symmetric
b. cylindrical symmetric
d. none of the above
a. experience a mild shock
b. experience a strong shock
c. get killed instantaneously
d. not be affected practically
d. may increase or decrease
The following questions have been compiled from a collection of questions submitted on Peer Wise by teacher candidates as part of the SYLLBUS of PHYSICS.
An electric field strength created by charge Q is measured to be 40 N/C at a distance of 0.2 m from the center of the charge. What is the new field strength when the distance from the center of Q is changed to 0.4 m away with twice the charge of Q?
A. 10 N/C
B. 20 N/C
C. 40 N/C
D. 80 N/C
Justification: Let the electric field strength be denoted by 𝐸𝐸. The magnitude of the electric field strength (𝐸𝐸) is defined as the force (𝐹𝐹) per charge (𝑞𝑞) on the source charge (𝑄𝑄). In other words, 𝐸𝐸 = 𝐹𝐹/𝑞𝑞 , where 𝐹𝐹 = 𝑘𝑘𝑘𝑘𝑘𝑘 𝑑𝑑2 is the electric force given by Coulomb’s law, k is the Coulomb’s law constant (𝑘𝑘 = 9.0 × 109 𝑁𝑁 𝑚𝑚2 𝐶𝐶2 ), and d is the distance
between the centers of 𝑞𝑞 and 𝑄𝑄.
So we need to use the expression, 𝐸𝐸 = 𝑘𝑘𝑘𝑘𝑘𝑘
𝑞𝑞𝑑𝑑2. Simplifying this expression gives, 𝐸𝐸 = 𝑘𝑘𝑘𝑘/𝑑𝑑2.
In our case, since 𝑄𝑄 and 𝑑𝑑 are doubled, the new field strength is 𝐸𝐸𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑘𝑘 (2𝑄𝑄) (2𝑑𝑑)2 , which can be simplified to get 𝐸𝐸𝑛𝑛𝑛𝑛𝑛𝑛 = 24 × 𝑘𝑘𝑘𝑘 𝑑𝑑2 = 12 𝐸𝐸.
Thus, the new field strength is 𝐸𝐸𝑛𝑛𝑛𝑛𝑛𝑛 = 12 𝐸𝐸 = 12 × 40 𝑁𝑁/𝐶𝐶 = 20𝑁𝑁/𝐶𝐶.
Despite doubling the charge from 𝑄𝑄 to 2𝑄𝑄 and the distance from 𝑑𝑑 to 2𝑑𝑑, our field strength 𝐸𝐸 decreased by half.
Finally, note that the expression for electric field strength illustrates an inverse square relationship between the electric field strength and the distance, 𝐸𝐸 ∝ 1 𝑑𝑑2.
Two point charges (C1 and C2) are fixed as shown in the setup below. Now consider a third test charge with charge -q that you can place anywhere you want in regions A, B, C, or D. In which region could you place the test charge so that the net force on the test charge is zero?
A. Region A
B. Region B
C. Region C
D. Region D
Answer: D – Somewhere in region D.
Justification: With the test charge and C1 being negative, there is a repulsive force on the test charge to the right. From C2, there is an attractive force on the test charge to the left. By referring to Coulomb`s law (𝐹𝐹 = 𝑘𝑘𝑞𝑞1𝑞𝑞2 𝑟𝑟2 ), we know that the force from C1 is being divided by a larger r so that the repulsive force between C1 and the test charge becomes smaller. However, the force from C2 and the test charge is being caused by a smaller magnitude of charge so that the attractive force between C2 and the test charge becomes smaller. At some point in region D, these two effects cancel out and there would be no net force on the test charge.
In region A, the net repulsive force from C1 would be much greater in strength than the attractive force from C2. This is because the C2 charge is greater than the C1 charge, and the test charge is much closer to C2. Therefore the net force would always be to the left (the test charge would be repelled away to the left).
In regions B and C, there would be a net repulsive force on the test charge from C2 to the right, as well as a net attractive force from C1 to the right as well. No matter where you placed the test charge in this region, it would always be pushed to the right.
In each of the four scenarios listed below, the two charges remain fixed in place as shown. Rank the electric potential energies of the four systems from the greatest to the least.
A. B = D > C > A
B. C > B > A > D
C. C > B = D > A
D. D > A = B > C
E. A > C > B = D
Justification: Recall that electric potential energy depends on two types of quantities: 1) electric charge (a property of the object experiencing the electrical field) and 2) the distance from the source (the location within the electric field). Somewhat similar to the gravitational potential energy, the electric potential energy is inversely proportional to 𝑟𝑟. The electric potential energy, 𝐸𝐸𝑃𝑃, is given by 𝐸𝐸𝑃𝑃 = 𝑘𝑘 𝑞𝑞1 𝑞𝑞2 𝑟𝑟 , where 𝑘𝑘 is the Coulomb’s law constant, 𝑞𝑞1 and 𝑞𝑞2 are point charges, and 𝑟𝑟 is the distance between
the two point charges. Note that 𝐸𝐸𝑃𝑃 is related to the electric force, 𝐹𝐹, given by Coulomb’s law. That is, 𝐸𝐸𝑃𝑃 = 𝐹𝐹 × 𝑟𝑟, where 𝐹𝐹 = 𝑘𝑘 𝑞𝑞1𝑞𝑞2 𝑟𝑟2 .
For system A: 𝐸𝐸𝑃𝑃 = 𝑘𝑘 4𝑞𝑞 × 𝑞𝑞 𝑑𝑑 = 4𝑘𝑘 𝑞𝑞2 𝑑𝑑
For system B: 𝐸𝐸𝑃𝑃 = 𝑘𝑘 3𝑞𝑞 × 3𝑞𝑞 𝑑𝑑 = 9𝑘𝑘 𝑞𝑞2 𝑑𝑑
For system C: 𝐸𝐸𝑃𝑃 = 𝑘𝑘 2𝑞𝑞 ×10𝑞𝑞 2𝑑𝑑 = 10𝑘𝑘 𝑞𝑞2 𝑑𝑑
For system D: 𝐸𝐸𝑃𝑃 = 𝑘𝑘 𝑞𝑞 ×𝑞𝑞 𝑑𝑑/3 = 3𝑘𝑘 𝑞𝑞2 𝑑𝑑
Since 𝑘𝑘 𝑞𝑞2 𝑑𝑑 is common to all of the above expressions, we note that the numerical coefficients determine the rank of the electric potential energies (i.e. 10 > 9 > 4 > 3). Thus B is the correct answer.
In each of the four scenarios listed below, the two charges remain fixed in place as shown. Rank the forces acting between the two charges from the greatest to the least.
A. C > B > A > D
B. C > B = D > A
C. B = D > C > A
D. B = D > A > C
E. A > C > B = D
Justification: Recall that the electric force is a fundamental force of the universe that exists between all charged particles. For example, the electric force is responsible for chemical bonds. The strength of the electric force between any two charged objects depends on the amount of charge that each object contains and also on the distance between the two charges. From Coulomb’s law, we know that the electric force is given by 𝐹𝐹 = 𝑘𝑘 𝑞𝑞1𝑞𝑞2 𝑟𝑟2 , where 𝑘𝑘 is the Coulomb’s law constant, 𝑞𝑞1 and 𝑞𝑞2 are point charges, and 𝑟𝑟 is the distance between the two point charges.
Note that 𝐹𝐹 is proportional to the amount of charge and also inversely proportional to the square of the distance between the charges.
For system A: 𝐹𝐹 = 𝑘𝑘 4𝑞𝑞 × 𝑞𝑞 𝑑𝑑2 = 4𝑘𝑘 𝑞𝑞2 𝑑𝑑2
For system B: 𝐹𝐹 = 𝑘𝑘 3𝑞𝑞 × 3𝑞𝑞 𝑑𝑑 = 9𝑘𝑘 𝑞𝑞2 𝑑𝑑2
For system C: 𝐹𝐹 = 𝑘𝑘 2𝑞𝑞 ×10𝑞𝑞 (2𝑑𝑑)2 = 5𝑘𝑘 𝑞𝑞2 𝑑𝑑2
For system D: 𝐹𝐹 = 𝑘𝑘 𝑞𝑞 ×𝑞𝑞 (𝑑𝑑/3)2 = 9𝑘𝑘 𝑞𝑞2 𝑑𝑑2
Since 𝑘𝑘 𝑞𝑞2 𝑑𝑑2 is common to all of the above expressions, we note that the numerical coefficients determine the rank of the electric forces (i.e. 9 = 9 > 5 > 4). Thus C is the correct answer.
Given the following electric field diagrams:
A. (a,b,c) = (-q, +q, +q) B. (a,b,c) = (+q, q, -q)
C. (a,b,c) = (+q, -q, -2q) D. (a,b,c) = (-q, +q, +2q)
E. (a,b,c) = (+2q, -2q, -q)
What are the respective charges of the yellow particles shown in diagrams (a), (b), and (c)?
Justification: Recall that the direction of an electric field is defined asthe direction that a positive test charge would be pushed when placed in the electric field. The electric field direction of a positively charged object is always directed away from the object. And also, the electric field direction of a negatively charged object is directed towards the
Since the field direction is directed away from (a) but towards (b) and (c), we know that the relative charges of (a,b,c) = (+,-,-) Note that the field lines allow us to not only visualize the direction of the electric field, but also to qualitatively get the magnitude of the field through the density of the field lines. From (a), (b), and (c), we can see that the density of the electric field lines in (c) is twice that of (a) or (b). We would expect the magnitude of the charge in (c) to also be twice as strong as (a) or (b). Thus, the answer choice C is correct.
Below is a diagram of a charged object (conductor) at electrostatic equilibrium. Points A, B, and D are on the surface of the object, whereas point C is located inside the object.
Rank the strength of the electric field at points A, B, C, and D from strongest to weakest.
A. B > D > A > C
B. B > D > C > A
C. D > B > C > A
D. D > B > A > C
E. A > B > D > C
Justification: We need to understand the concept of the electric field being zero inside of a closed conducting surface of an object, which was demonstrated by Michael Faraday in the 19th century. Suppose to the contrary, if an electric field were to exist below the surface of the conductor, then the electric field would exert a force on electrons present there. This implies that electrons would be in motion. However, the assumption that we made was that for objects at electrostatic equilibrium, charged particles are not in motion. So if charged particles are in motion, then the object is not in electrostatic equilibrium. Thus, if we assume that the conductor is at electrostatic equilibrium, then the net force on the electrons within the conductor is zero. So at point C, the electric field is zero.
FAQs on Electrostatic
What is the formula for electrostatic?
Calculate the electrostatic force using the formula: F = K[q1 x q2]/D^2 where K is coulombs constant, which is equal to 9 x 10^9 Nm^2/C^2. The unit for K is newtons square meters per square coulombs.
What are the important topics in electrostatics?
The topics covered here are:
Potential due to a point charge,
Potential due to a system of charges,
Relation between field and potential,
What is an example of electrostatic?
Examples of electrostatic forces: When we run a piece of paper with the oil in our head with the help of a comb produces electrostatic force. Balloons get attracted to another balloon when one of them are rubbed with hair.