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In this Class 12 Physics topic from Chapter 1, Electric Charges and Fields, students learn the basic nature of electric charge and the law of conservation of charge. They understand that charge can neither be created nor destroyed, but may be transferred between bodies through processes such as rubbing, contact, or induction. The topic also builds a foundation for analysing charged systems and applying charge conservation while studying electric fields and related phenomena.
TOPIC PRACTICE
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Hard · Level 2View options
Addition with signs, conservation, and checking integral multiples of elementary charge
Magnitude-only addition, direction, and colour
Temperature, mass, and smell
Only attraction, only repulsion, and no calculation
Hard · Level 2View options
Total charge is conserved, but distribution and the number of electrons on a body can change
Total charge always becomes zero
Charge never transfers
A positive body has no electrons
Hard · Level 2View options
In an open system the chosen object's charge always remains constant
Total charge remains constant in an isolated system while charge may cross the chosen boundary in an open system
Charge is destroyed in an isolated system
Charge has no existence in an open system
Hard · Level 2View options
Negative two elementary charges
Positive two elementary charges
Negative sixteen elementary charges
Zero
Hard · Level 2View options
Positive six elementary charges
Positive twelve elementary charges
Negative six elementary charges
Positive eleven elementary charges
Hard · Level 2View options
Because each would get half an elementary charge, which is not possible for a free sphere
Because total charge becomes zero
Because positive charge changes into negative
Because electrons disappear
Hard · Level 2View options
First decide the isolated system, then add all charges with signs
First remove all negative signs
First write the final charge by guess
First take only the largest charge
Hard · Level 2View options
Near the surface the field is nearly uniform and perpendicular to the surface
Because the field inside the conductor is maximum
Because lines flow along the surface
Because charge cannot remain on a conductor
Hard · Level 2View options
Because higher curvature there can produce a stronger external field
Because charge is destroyed at sharp parts
Because the field inside the conductor is maximum
Because charge cannot remain on flat parts
Hard · Level 2View options
Because the fields of opposite charges partially cancel
Because a dipole has no electric field
Because both charges always produce fields in the same direction
Because distance has no effect
Hard · Level 2View options
Because field direction is not defined there
Because potential must be zero there
Because there must be no charge there
Because field lines are only straight
Question 1HardLevel 2
Which combined use of charge properties is most useful in numerical questions?
Correct answer: A
Reliable charge calculations require several connected checks. First, use signed algebraic addition to obtain the net charge rather than adding magnitudes blindly. Second, apply conservation to compare the total charge before and after transfer or contact. Third, where microscopic charge is relevant, check that the result is an integral multiple of the elementary charge e. Option A combines all three valid tools; the other choices contain irrelevant or incomplete ideas.
What is the most correct conclusion for difficult questions based on electric charge and conservation?
Correct answer: A
The central rule is conservation of the total charge of an isolated system, not conservation of the charge on every individual body. During rubbing, contact, or induction, electrons can move between bodies, changing their charge distribution and the number of electrons on each body. Their total algebraic charge, however, remains constant. Thus A is the most complete conclusion. B and C deny the rule’s actual meaning, while D wrongly claims that a positive body contains no electrons.
Which statement correctly gives the difference between an isolated system and an open system for charge conservation?
Correct answer: B
Charge conservation applies to the total charge of a suitably isolated system. No charge crosses its boundary, so the algebraic total remains constant. In an open system, charge can enter or leave across the selected boundary; consequently, the charge contained inside that boundary may change even though charge is conserved in the larger system. Thus option B states the correct distinction. The other options deny conservation or make absolute claims.
In an isolated system, three bodies have charges positive seven elementary charges, negative eleven elementary charges, and positive two elementary charges. What will be the total charge after mutual contact?
Correct answer: A
For an isolated system, mutual contact may redistribute charge among the bodies, but it cannot change the algebraic total. Add the signed charges: (+7e) + (−11e) + (+2e) = (7 − 11 + 2)e = −2e. Therefore the total charge after contact is negative two elementary charges, option A. Options B and C result from sign or arithmetic errors, while D incorrectly assumes complete cancellation.
Two identical conducting spheres carry positive seventeen elementary charges and negative five elementary charges. What charge will each have after touching and separation?
Correct answer: A
First find the conserved total charge: (+17e) + (−5e) = +12e. Because the spheres are identical conductors, touching allows charge to redistribute until both spheres have equal charge. The total is therefore divided equally: +12e/2 = +6e on each sphere. Hence option A is correct. Option B is the total charge mistakenly assigned to each sphere, while C has the wrong sign and D uses an incorrect average.
Two identical conducting spheres have positive one elementary charge and zero charge. Why does equal sharing create a difficulty?
Correct answer: A
The total initial charge is +e. A classical equal-sharing calculation for identical spheres would assign +e/2 to each sphere. However, the net charge of an isolated free body is quantized in integral multiples of e, so two independently observable spheres cannot each retain exactly half an elementary charge. This illustrates the microscopic limitation of the macroscopic equal-sharing model; total charge itself remains +e.
What is the safest method in difficult numerical problems on charge conservation?
Correct answer: A
The reliable method follows the conservation equation Q_initial = Q_final for a clearly defined isolated system. First decide which bodies belong to the system; then list every initial and final charge with its positive or negative sign. Add them algebraically and solve for the unknown while keeping the two totals equal. Thus A is correct. Removing signs, guessing, or selecting only the largest charge discards essential physical information.
Near the surface of a large plane conductor, why are field lines approximately parallel and perpendicular to the surface?
Correct answer: A
In electrostatic equilibrium, the electric field inside a conductor is zero and the field immediately outside its surface has no tangential component. If a tangential component existed, free charges would move along the surface, contradicting equilibrium; therefore the external field is normal, or perpendicular, to the conductor. For a large plane surface, edge effects and curvature are negligible, so the field is approximately uniform over a small nearby region and the lines are parallel to one another. Thus option A is correct.
Why is charge denser near sharp parts of an irregular conductor in electrostatic condition?
Correct answer: A
For an electrostatic conductor, the tangential electric field at the surface is zero, but the normal field just outside is related to surface charge density by E = σ/ε₀. At a sharp region, the radius of curvature is small, so charge crowds there and σ becomes large; consequently the external field is stronger. Therefore A is correct. Charge is not destroyed, the internal field remains zero, and flat regions can also carry charge.
Why does the far field of an electric dipole decrease faster with distance than the field of a point charge?
Correct answer: A
A point charge has a monopole field that varies as 1/r². An electric dipole contains equal and opposite charges, so at a far point their leading 1/r² contributions nearly cancel. The remaining dipole term varies as 1/r³, for example Eaxial ≈ 2p/(4πε₀r³). Hence the dipole field decreases faster. Option A identifies the cancellation; B, C, and D contradict the dipole field law.
At a point electric field is zero, but nearby field lines curve around it. Why should a field line not be drawn through the zero point?
Correct answer: A
An electric field line is defined so that its tangent at any point gives the direction of the electric field at that point. At a zero-field point, the vector has zero magnitude and therefore no unique direction. Although field lines in the surrounding region may curve around the point, no definite tangent can be assigned exactly there. Zero field does not necessarily mean zero potential or absence of charge, and field lines need not be straight.
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