Electric Field and Electric Field Lines

Correct answer: A. It becomes double

Explanation: 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.

Correct answer: A. It increases

Explanation: 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.

Correct answer: A. The diagram is wrong because tangential field should not exist on the surface

Explanation: 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.

Correct answer: A. 13 newton per coulomb

Explanation: 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.

Correct answer: A. The resultant field will no longer be zero

Explanation: 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.

Correct answer: A. It is a qualitative comparison, not a count of real lines

Explanation: 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.

Correct answer: A. Because it would imply two directions at a point and a closed path

Explanation: 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.

Correct answer: A. Along the perpendicular bisector away from the two charges

Explanation: 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.

Correct answer: A. Magnitude may be nearly same but direction is changing

Explanation: 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.

Correct answer: A. No, fields due to source charges can cancel each other

Explanation: 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.

Correct answer: A. 100 newton per coulomb

Explanation: 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.

Correct answer: A. By adding the magnitudes

Explanation: 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.

Correct answer: A. In the direction of the larger field

Explanation: 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.

Correct answer: A. The net charge is negative

Explanation: 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.

Correct answer: A. More dense and larger in number going outward

Explanation: 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.

Correct answer: A. By the tangent drawn at that point

Explanation: 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.

Correct answer: A. Direction of lines reverses but density remains the same

Explanation: 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.

Correct answer: A. From infinity

Explanation: 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.

Correct answer: A. Positive

Explanation: 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.

Correct answer: A. Net charge may be zero

Explanation: 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.

Correct answer: A. The field is not uniform near the edges

Explanation: 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.

Correct answer: A. Direction is same but magnitude is different

Explanation: 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.

Correct answer: A. Same at every point

Explanation: 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.

Correct answer: A. Because directions are different and radial

Explanation: 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.

Correct answer: A. No, the field is determined by source charges and position

Explanation: 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.