Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
In this Class 12 Physics topic from Chapter 1, Electric Charges and Fields, students learn how electric charges produce an electric field and how the field is represented using electric field lines. The topic explains field strength, direction, the role of a test charge, and the principle of superposition for multiple charges. Students also study the properties, patterns, and relative density of field lines, including their use in understanding isolated charges and electric dipoles.
TOPIC PRACTICE
Quiz this set
Up to 25 questions from this page. Select your focus, then start.
25 questions
Choose questions
Medium · Level 9View options
It becomes double
It becomes half
It becomes four times
It remains unchanged
Medium · Level 9View options
It increases
It decreases
It is always zero
It depends only on a change of direction
Medium · Level 9View options
The diagram is wrong because tangential field should not exist on the surface
The diagram is correct because field is always parallel to a conductor
The diagram is correct because field is maximum inside a conductor
No conclusion can be drawn
Medium · Level 9View options
13 newton per coulomb
17 newton per coulomb
7 newton per coulomb
60 newton per coulomb
Medium · Level 9View options
The resultant field will no longer be zero
The resultant field must still be zero
The resultant field becomes infinite
Both fields become scalar quantities
Medium · Level 9View options
It is a qualitative comparison, not a count of real lines
It is true only for negative charges
It is true only for zero field
It cannot show field direction
Medium · Level 9View options
Because it would imply two directions at a point and a closed path
Because every field line must be closed
Because positive charge produces no field
Because a field line is drawn only in a plane
Medium · Level 9View options
Along the perpendicular bisector away from the two charges
Along the line joining the charges
Towards one charge
No direction because the field is zero
Medium · Level 9View options
Magnitude may be nearly same but direction is changing
The field is completely uniform
The field is necessarily zero
The field has no source
Medium · Level 9View options
No, fields due to source charges can cancel each other
Yes, zero force means no charge exists
Yes, because field is made only by one charge
No, because positive test charge never feels force
Medium · Level 9View options
100 newton per coulomb
300 newton per coulomb
2700 newton per coulomb
8100 newton per coulomb
Medium · Level 9View options
By adding the magnitudes
By subtracting the smaller from the larger
By always taking zero
By taking only the smaller field
Medium · Level 9View options
In the direction of the larger field
In the direction of the smaller field
In any direction
Always perpendicular
Medium · Level 9View options
The net charge is negative
The net charge is positive
The net charge is zero
There is no charge in the diagram
Medium · Level 9View options
More dense and larger in number going outward
Less dense and going inward
Forming closed circles
Ending on the other charge
Medium · Level 9View options
By the tangent drawn at that point
By the total length of the line
By the mass of the charge
By the edge of the diagram
Medium · Level 9View options
Direction of lines reverses but density remains the same
Lines disappear
Density becomes four times
Lines become closed circles
Medium · Level 9View options
From infinity
From the same negative charge
From inside a conductor
They are not real, so they have no direction
Medium · Level 9View options
Positive
Negative
Zero
Cannot be determined
Medium · Level 9View options
Net charge may be zero
Net charge is definitely positive
Net charge is definitely negative
Electric field inside must be zero
Medium · Level 9View options
The field is not uniform near the edges
Charge disappears at the edges
Field lines become closed at the edges
The unit of field changes at the edges
Medium · Level 9View options
Direction is same but magnitude is different
Magnitude is same but direction is different
Field is zero at both points
Field is completely identical at both points
Medium · Level 9View options
Same at every point
Greater on the upper part
Zero at only one point
Direction same everywhere
Medium · Level 9View options
Because directions are different and radial
Because magnitudes are actually not same
Because a point charge produces no field
Because electric field is not a vector
Medium · Level 9View options
No, the field is determined by source charges and position
Yes, the field is always proportional to the test charge
Yes, a field cannot exist without a large test charge
No, because the field has no source
Question 1MediumLevel 9
If the distance from a point charge is halved and the charge magnitude is also halved, how will the electric field change?
Correct answer: A
For a point charge, the field magnitude is E = k|q|/r². Let the original values be q and r, so E = kq/r². After the changes, q' = q/2 and r' = r/2. Hence E' = k(q/2)/(r/2)² = (kq/2)/(r²/4) = 2E. The charge reduction halves the field, while the distance reduction multiplies it by four, giving option A.
In a non-uniform electric field, field lines become closer in one direction. Moving in that direction, how does the field magnitude change?
Correct answer: A
The density of electric-field lines is used to represent the relative magnitude of the field: more lines crossing a given area indicate a stronger field. Therefore, when the lines become progressively closer in a direction, the line density increases and the field magnitude increases along that direction. This does not mean the field is zero, and the conclusion is not based merely on a change of direction. Hence option A is correct.
If field lines in a diagram touch a conductor while running along its surface, what conclusion is correct according to electrostatic equilibrium?
Correct answer: A
The governing condition is electrostatic equilibrium: free charges in a conductor rearrange until the tangential component of the electric field at its surface becomes zero. A field line drawn along the surface would represent a nonzero tangential component and would cause charges to move, contradicting equilibrium. Therefore, the diagram is incorrect, and valid electric field lines must meet a conductor’s surface normally. Option B reverses the correct condition, while C incorrectly describes the field inside a conductor.
At a point, fields of 5 newton per coulomb eastward and 12 newton per coulomb northward act. What is the magnitude of the resultant field?
Correct answer: A
The eastward and northward electric-field vectors are perpendicular, so their resultant magnitude follows the Pythagorean relation: E = √(E₁² + E₂²) = √(5² + 12²) N/C = √(25 + 144) N/C = √169 N/C = 13 N/C. Therefore option A is correct. The value 17 N/C is the simple arithmetic sum and would apply only to parallel fields in the same direction; 7 N/C is a difference, and 60 has no valid vector basis.
If two equal electric fields in opposite directions give zero resultant, what happens if the direction of one is slightly turned?
Correct answer: A
Exact cancellation of two vectors requires two simultaneous conditions: equal magnitudes and exactly opposite directions. Initially both conditions hold, so the resultant is zero. When one field is turned even slightly, the directions are no longer antiparallel; therefore the vector components do not cancel completely and a nonzero resultant remains. Its exact magnitude depends on the turning angle, but it is not automatically infinite or zero. The vector quantities also remain vectors, so D is incorrect.
The density of electric field lines indicates field magnitude. What is the most correct limitation of this statement?
Correct answer: A
Electric field lines are an illustrative representation, not physical strings or independent objects. In a properly scaled diagram, a greater density of lines indicates a stronger field and a lower density indicates a weaker field, so the idea is useful for qualitative comparison. However, the number of drawn lines is chosen by the diagram-maker and is not a literal count of field entities. The lines can also show direction through their tangent and arrows, so D is too strong.
In a field-line diagram, a line crosses itself and forms a closed shape. Why is this unacceptable for an electrostatic field?
Correct answer: A
The tangent to an electric field line gives the direction of the electric field at that point. If one line crossed itself, the same point would have two different tangents and therefore two field directions, which is impossible for a single-valued vector field. Moreover, electrostatic fields are conservative, so their field lines do not form closed loops. Thus option A is correct; the other choices contradict basic electrostatics.
For two equal positive charges, at a very far point on the perpendicular bisector, what is the approximate direction of the combined field?
Correct answer: A
At a point on the perpendicular bisector, the two charges are equally distant, so their field magnitudes are equal. The components parallel to the line joining the charges are opposite and cancel, while the components along the perpendicular bisector point in the same outward direction and add. The resultant is therefore along the perpendicular bisector, away from the charges. Option A is correct.
In a region, field lines are equally spaced but gradually curved. What conclusion follows?
Correct answer: A
The density or spacing of field lines is used qualitatively to represent field magnitude, so nearly equal spacing suggests that the magnitude may remain approximately constant. However, the tangent to each line gives the local field direction. Because the lines are curved, their tangents change from place to place, so the direction changes. The field is therefore not uniform, making option A correct.
A positive test charge experiences zero force at a point. Is it correct to say that there are no source charges around?
Correct answer: A
The force on a test charge is F = qE. For a nonzero positive test charge, zero force means that the net electric field at that point is zero. It does not prove that every source charge is absent. Fields produced by several charges can have equal magnitudes in opposite directions and cancel by superposition. Thus option A is correct; zero net field is not the same as no sources.
The field due to a point charge is 900 newton per coulomb at a distance of 0.2 metre. What will be the field at 0.6 metre?
Correct answer: A
For a point charge, the electric-field magnitude follows the inverse-square law, E = kQ/r². The new distance is 0.6/0.2 = 3 times the original distance, so the field becomes 3² = 9 times smaller. Therefore, E₂ = 900/9 = 100 N/C. Option B would result from inverse proportionality without the square; C and D incorrectly increase the field as distance increases.
At a point, electric field vectors due to two charges have the same direction. How is the magnitude of the net field found?
Correct answer: A
Electric field is a vector quantity, so both magnitude and direction must be considered. When two field vectors point in exactly the same direction, their resultant magnitude is the arithmetic sum: E_net = E₁ + E₂. Subtraction is appropriate for opposite directions, not for vectors that reinforce one another. The net field is not automatically zero and cannot be represented by only the smaller contribution. Therefore, option A is correct.
At a point, electric field vectors due to two charges are opposite and unequal in magnitude. What is the direction of the net field?
Correct answer: A
For two opposite electric-field vectors, the resultant magnitude is the difference between their magnitudes: E_net = |E₁ − E₂|. Because the magnitudes are unequal, complete cancellation cannot occur. The larger vector leaves an excess in its own direction, so the net field points in the direction of the larger field. The smaller direction would be correct only if it were larger, and perpendicularity does not arise for exactly opposite vectors.
In a field-line diagram, fewer lines start from positive charge and more lines terminate on negative charge. What does this indicate about net charge?
Correct answer: A
Electric field lines conventionally originate on positive charge and terminate on negative charge. The number of lines is used to represent the relative magnitude of charge. If more lines end on negative charge than leave positive charge, there is an excess of negative charge in the system, so the net charge is negative. A zero net charge would require balanced positive and negative line counts; a positive net charge would show the opposite imbalance.
A field-line diagram has two unequal positive charges. How should lines appear near the larger positive charge?
Correct answer: A
Field lines emerge outward from a positive charge, and their number represents the charge magnitude while their local spacing indicates field strength. Therefore, the larger positive charge must have more lines, and those lines should be more closely spaced near it, showing greater field intensity. Lines do not point inward toward a positive charge, form closed circles in electrostatics, or terminate on another like positive charge. Hence option A is correct.
If an electric field line is curved at a point, how is the field direction at that point decided?
Correct answer: A
An electric field line is defined so that its tangent at any point gives the instantaneous direction of the electric field there. For a curved line, the direction can change continuously from one point to another, so the entire curve cannot be used as one single direction. Draw a tangent at the specified point and use the arrow on that tangent to determine the direction. Thus option A is correct.
The sign of a point charge is changed while magnitude and distance remain the same. What changes in the field-line diagram?
Correct answer: A
For a point charge, the field magnitude is E = k|Q|/r², so it depends on the magnitude of charge and the distance, not on the sign. Changing a positive charge to a negative one therefore reverses the line direction: lines that emerged outward now terminate inward. Since |Q| and r are unchanged, the relative number and density of lines remain unchanged. The lines do not disappear, multiply fourfold, or become closed circles.
If field lines terminate on a negative charge and their number is more than the lines starting from positive sources, from where are the extra lines considered to come?
Correct answer: A
Electric field lines are drawn from positive charges toward negative charges. If a negative charge has more terminating lines than the lines supplied by positive charges, the additional lines are represented as originating at infinity and ending on that negative charge. This signifies a net negative charge in the considered system. They cannot originate from the same negative charge, and field lines still have a defined direction even though they are a visualization.
If more electric field lines leave a closed surface than enter it, what sign of net charge inside is indicated?
Correct answer: A
The net outward flux through a closed surface is related to the enclosed charge by Gauss’s law, Φ = Q enclosed divided by ε0. More field lines leaving than entering means the net flux is outward and positive. Therefore, the enclosed net charge is positive. A negative charge would produce a net inward pattern, while zero net charge would give balanced inward and outward flux.
If the same number of field lines enter and leave a closed surface, what conclusion about net charge inside is appropriate?
Correct answer: A
Equal numbers of field lines entering and leaving a closed surface represent zero net electric flux through that surface. By Gauss’s law, zero net flux means the algebraic net charge enclosed is zero, so option A is the appropriate conclusion. It does not mean that no charges are present: equal positive and negative charges may coexist, and the electric field can still be nonzero at points inside.
Between large parallel plates, field lines are nearly straight and equally spaced. Why may they bend near the edges?
Correct answer: A
The central region between two large, oppositely charged parallel plates is treated as a nearly uniform electric field, so the lines are straight and equally spaced there. Near an edge, the ideal infinite-plate geometry ends; charge distribution and field direction spread outward, producing fringing. The charge does not disappear, field lines do not close, and the unit of electric field remains unchanged.
If field-line direction is the same at two points but density is different, which statement about the fields is correct?
Correct answer: A
An electric field is a vector, so it is specified by both direction and magnitude. In a field-line diagram, the tangent to a line gives direction, whereas the relative density of lines represents magnitude. Thus, identical directions combined with different densities indicate equal direction but unequal magnitudes. The field is not necessarily zero, and the two vectors are not completely identical.
In the field of a point charge, all points on a spherical surface are at the same distance from the charge. How is the field magnitude on that surface?
Correct answer: A
The electric-field magnitude due to a point charge is E = (1/4πε0)|q|/r². On a spherical surface centered on that charge, r has the same value at every point, and q is fixed; therefore the magnitude is identical everywhere on the surface. However, the direction is radial and changes from point to point. Option D describes direction, not magnitude, and is therefore not correct.
Around a point charge, field magnitude is same at equal distance. Why are the field vectors still not identical?
Correct answer: A
Electric field is a vector quantity, so two field vectors are identical only when both their magnitudes and directions are identical. For a point charge, E = k|q|/r² gives the same magnitude at every point having the same distance r. Nevertheless, the field direction is radial: it points away from a positive charge or toward a negative charge, and radial directions differ around the sphere. Hence option A is correct.
If the electric field at a point is very large, does it mean that a very large test charge is placed there?
Correct answer: A
The electric field is a property of the source-charge arrangement at a point, defined as E = F/q₀ for a sufficiently small positive test charge. The test charge measures the field but does not create its value. A large field can result from large source charges or a small distance from them. Therefore, option A is correct; B and C incorrectly assign the field to the test charge, while D denies the source of the field.
Google Analytics helps us understand site usage. Google may send limited cookie-free signals before your choice. The Live Visitors widget operates independently of this analytics choice; see the privacy policy for its provider and fallback details. Essential site features work without analytics cookies. You can change your choice later in Privacy choices. Privacy policy