Electric Field and Electric Field Lines

Electric field at a point is defined as E = F/q for a test charge that is sufficiently small and does not disturb the source-charge configuration. A very large positive test charge can exert an appreciable force on the source charges, shifting them or changing their distribution. The measured field would then no longer represent the original field. Therefore option A is correct; the other choices make unsupported absolute claims about zero force or force direction.

The density of electric-field lines is used qualitatively to represent field strength: closely spaced lines indicate a strong field. A uniform field has the same magnitude and direction throughout a region, whereas rapid local change in line direction indicates that the field is non-uniform. Combining both observations gives a strong but non-uniform field, so option A is correct. B contradicts both clues, C contradicts the high density, and D ignores the meaning of the tangent direction.

Let each field have magnitude E and let the angle between them be 60°. Vector addition gives R = √(E² + E² + 2E² cos60°) = √3E. Since √3E is greater than E but less than 2E, the resultant is greater than either one field. Therefore option B is correct. It cannot exceed the arithmetic sum, so C is false; A is too weak as the intended comparison, and D is impossible.

Electric field-line arrows indicate the direction of the electric field, namely the direction of force on a positive test charge. In a small smooth region where nearby lines are nearly parallel, the field direction should vary continuously, not reverse suddenly from one nearby line to another without a source, sink, or zero-field boundary. Thus option A identifies the diagram error. Parallel lines are valid, arrows are necessary, and the field need not be zero.

The governing concept is the direction of electric field lines and Gauss’s law. Field lines originate from positive charges and terminate on negative charges or extend to infinity. Therefore, lines emerging outward through a closed surface are consistent with a net positive charge enclosed by that surface. Option B would produce inward lines, C is not implied by line crossing, and D is false because field lines may pass through an imaginary closed surface.

A valid electric field line cannot stop abruptly at an ordinary point in space. Field lines begin on positive charges and end on negative charges, or they may continue to infinity when no terminating negative charge is present in the displayed region. Thus an unexplained sudden ending indicates an incomplete or incorrect diagram. Option B contradicts the field-line convention, C does not follow from an ending, and D makes an unsupported conclusion.

The governing concept is vector addition. Since the two electric-field vectors are perpendicular, their resultant magnitude is found using the Pythagorean relation: E = √(E₁² + E₂²) = √(6² + 8²) = √(36 + 64) = √100 = 10 N/C. Thus option C is correct. Fourteen is ordinary scalar addition, two is the difference, and forty-eight is the product, none of which applies here.

The governing equation is F = qE, where E is the net electric field at the location. Since E = 0, multiplication by either a positive or negative finite test charge gives F = q × 0 = 0. Changing the sign would reverse a nonzero force, but it cannot create force where the net field is zero. Thus option A is correct; the other choices assume a direction or infinite value without any field.

The governing concept is vector superposition and the direction of field lines. At the midpoint, the distances from the equal positive and negative charges are equal, so the two field magnitudes are equal. However, the positive charge produces a field away from itself, while the negative charge produces a field toward itself. Between them, both vectors point from positive to negative, so they add rather than cancel. Hence option A is correct.

Electric field obeys vector addition. The north component is 4 N/C and the east component is 3 N/C, so the resultant has both positive northward and eastward components. Its magnitude would be sqrt(4^2 + 3^2) = 5 N/C, and its direction satisfies tan theta = 3/4 measured east of north. Thus it lies between north and east, closer to north. Option A is correct; B and C omit one component, while D has opposite signs.

At the midpoint, the electric fields produced by the two equal positive charges have equal magnitudes and opposite directions, so their vector sum is zero. However, electric potential is scalar. If each charge is +Q and its distance from the midpoint is r, the total potential is V = kQ/r + kQ/r = 2kQ/r, which is positive and nonzero. Therefore option A is correct; B has the wrong sign, C confuses field cancellation with potential cancellation, and D treats a scalar as directional.

An electric field line represents the instantaneous direction of the electric force on a positive test charge; it is not automatically the particle's actual trajectory. The trajectory depends on Newton's second law, F = qE, as well as the initial velocity, mass, and how the field changes with position. Only in special conditions, such as release from rest in a suitably uniform or straight field, can the path coincide with a field line. Thus A is correct.

Electrostatic fields are conservative, so their circulation around any closed path is zero: ∮E·dl = 0. Consequently, electrostatic field lines do not form closed loops; they originate on positive charges and terminate on negative charges or extend to infinity. Closed loops are associated with magnetic field lines, not electrostatic ones. Hence option A gives the appropriate physical description. B denies the vector nature of the field, while C and D reverse or misuse the rule.

In a field-line diagram, the number of lines per unit area represents the relative magnitude of the electric field. When the lines spread apart, fewer lines pass through the same area, so their density decreases and the field is weaker in that later region. Therefore A is correct. Closer spacing would indicate a stronger field; spreading does not mean lines intersect, and it certainly does not imply zero field everywhere.

For a point charge, the electric-field magnitude follows E = kQ/r². Let the distances be r_A and r_B. Given E_A = 4E_B, we have kQ/r_A² = 4kQ/r_B², so r_B² = 4r_A² and therefore r_B = 2r_A. Thus point A is half as far from the charge as point B, making A correct. The inverse-square relation is why the distance ratio is not four or two in the same direction as the field ratio.

The point-charge relation is E = kQ/r². Tripling the source charge multiplies E by 3. Halving the distance changes r to r/2, so the inverse-square factor multiplies E by 1/(1/2)² = 4. Applying both independent changes gives E' / E = 3 × 4 = 12. Hence the field becomes twelve times its original value, so A is correct; the other options omit or mishandle one of the two factors.

The electric force on a charge is F_e = qE. For a negative charge, q is negative, so its electric force is opposite to the eastward field: it acts westward. To keep the charge at rest, the net force must be zero, so the external force must have equal magnitude and act opposite to the electric force, namely eastward. Therefore A is correct; westward would reinforce the electric force.

For a positive charge, the electric force F_e = qE is in the same direction as the electric field, so an eastward field pushes the charge eastward. Rest requires zero resultant force. Consequently, the external force must be equal in magnitude and opposite in direction, namely westward. Option A is correct. An eastward external force would add to the electric force, while northward or downward forces would not cancel the eastward component.

Electric field lines are a representation of the field produced by the source charges, not a count of lines created by the test charge. An ideal test charge is sufficiently small that it does not disturb the source charge distribution. Doubling that still-negligible test charge changes the force on it, since F = qE, but it does not change E or the source field-line diagram. Thus A is correct; lines do not automatically double, close, or intersect.

The governing relation is F = qE, where q is the test charge and E is the electric field. If the field changes to 2E and the charge changes to q/2, the new force is F' = (q/2)(2E) = qE = F. Therefore, the two changes cancel exactly and the force remains unchanged. Option B considers only the field change, while option C considers only the charge change.

Electric field direction is defined as the force direction on a positive test charge. A negative charge experiences force in the direction opposite to the field because F = qE and q is negative. Thus, when the field points downward, the electric force and corresponding acceleration of a negative charge can point upward, assuming electric force is the relevant net force. Hence option A is correct.

For a positive charge, the electric force has the same direction as the electric field, so it is rightward. In a field-line diagram, greater line density represents greater field magnitude. Since the lines become denser toward the right, E and therefore F = qE increase in that direction for a fixed positive charge. The field is non-uniform, so options C and D are inconsistent with the changing density.

An electric field is uniform only when its magnitude and direction remain constant throughout the region. Near the edges of parallel plates, the lines curve outward, so their direction changes; their spacing also changes, indicating a change in magnitude. This boundary effect is called fringing. The field is approximately uniform only in the central region, so option A is correct rather than claiming zero field or intersecting lines.

Electric field lines are imaginary graphical representations, not physical objects that can be counted in space. However, when diagrams use the same drawing convention, the relative number of lines emerging from or entering charges can represent relative charge magnitudes. The relationship is therefore qualitative or comparative, not an absolute unit conversion. Hence option A is correct; options B and C treat symbolic lines as physical quantities.

For a charge q in an electric field, the force is F = qE. Since q is positive, the force has the same direction as the electric field arrows. Field-line density represents field strength; moving from dense to sparse lines indicates that E decreases, so the force may also decrease as the charge moves forward. Hence option A is correct. The other options reverse the direction or ignore non-uniformity.