Electric Field and Electric Field Lines

Correct answer: A. Southward

Explanation: The tangent to an electric field line gives the direction of the electric field at that point. The force on a charge is F = qE. For a negative charge, q is negative, so the force direction is opposite to E. Since the field points north, the force points south; therefore option A is correct. A northward force would apply to a positive charge, not a negative one.

Correct answer: A. 4.5 newton per coulomb

Explanation: For a point charge, the electric-field magnitude follows E = k|Q|/r². Halving Q multiplies E by 1/2, while doubling r multiplies it by 1/2² = 1/4. The combined factor is 1/8. Therefore E_new = 36 × 1/8 = 4.5 N/C, so option A is correct. The other values result from ignoring one change or treating distance as a direct rather than squared factor.

Correct answer: A. The electric field remains the same

Explanation: Electric field at a point is defined as the force per unit positive test charge, E = F/q, produced by the source charges. The test charge is only a small probe and does not create the field being measured in the ideal definition. If the source charges and their positions are unchanged, the field remains unchanged. Thus option A is correct; removing the probe does not make the field zero or reverse it.

Correct answer: A. No, zero field does not necessarily mean zero potential

Explanation: Electric field and electric potential are related but distinct quantities. The field is the negative spatial rate of change of potential, E = -dV/dr. A point can have zero field because opposing field vectors cancel, while the scalar potentials add to a nonzero value. For example, the midpoint of equal like charges has zero field but positive potential. Therefore option A is correct.

Correct answer: A. The charge with the larger magnitude

Explanation: The number of field lines is a qualitative representation of charge magnitude, not an exact physical count. Therefore, the charge having the larger absolute value is drawn with more lines. Lines emerge from a positive charge and terminate on a negative charge, but the sign determines the end or start, while magnitude determines the relative number.

Correct answer: A. The net charge is positive

Explanation: Electric field lines conventionally originate on positive charge and terminate on negative charge or at infinity. If some lines leave the positive charge without ending on available negative charge, the positive charge exceeds the total negative charge in magnitude. Thus the system has positive net charge, and the unpaired lines extend to infinity.

Correct answer: A. The field is non-uniform and stronger in the dense region

Explanation: The density of field lines represents the relative magnitude of the electric field: closer lines indicate a stronger field, while wider spacing indicates a weaker field. Since the spacing changes with position, the field is non-uniform. Option B would require nearly constant spacing; options C and D contradict the meaning of line density and the presence of a source.

Correct answer: A. Some lines from the positive charge will go to infinity

Explanation: Field lines originate on positive charges and terminate on negative charges or at infinity. Let the positive charge magnitude be larger than the negative charge magnitude. The negative charge can absorb only a corresponding number of lines, so the surplus lines must leave the positive charge and extend to infinity. This also signals a positive net charge.

Correct answer: A. From infinity and ending on the negative charge

Explanation: Electric field lines originate on positive charges and terminate on negative charges. If the negative charge has greater magnitude, it must receive more lines than the positive charge can supply. The additional lines are therefore represented as originating at infinity and ending on the negative charge. Option B is impossible because field lines do not end on a positive charge, while C reverses the direction and D incorrectly treats electric field lines as closed loops.

Correct answer: A. Twice

Explanation: For a point charge, the electric-field magnitude is E = k|q|/r², so E is inversely proportional to the square of distance. Let the original field be E₁ = 100 and the new field be E₂ = 25. Then E₂/E₁ = 1/4 = (r₁/r₂)², giving r₂/r₁ = 2. Thus the required distance is twice the original distance. Four times would reduce the field by sixteen, not four.

Correct answer: A. Because the net field at each point is the vector sum of the fields due to both charges

Explanation: A single point charge produces a field with a simple radial direction, so its field lines are straight rays. For two charges, the electric field at any point is the vector sum of the fields produced by both charges. Since the relative directions and magnitudes change from point to point, the resultant direction also changes continuously. Field lines follow this changing direction and can therefore be curved. They are not arbitrary drawings and are not restricted to negative charges.

Correct answer: A. In electrostatic equilibrium, lines must be perpendicular to the surface

Explanation: In electrostatic equilibrium, free charges in a conductor have no net motion. If the external electric field had a tangential component along the surface, it would exert a force on these free charges and make them move. Charges redistribute until the tangential component becomes zero. Consequently, the electric field just outside the conductor is normal to the surface, so oblique field lines indicate an error. A field can exist outside a charged conductor.

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

Explanation: Take a point on the perpendicular bisector of two equal positive charges. Its distances from the charges are equal, so the two field magnitudes are equal. The components parallel to the line joining the charges are opposite and cancel by symmetry. The components along the perpendicular bisector point away from each positive charge and therefore add. The resultant field is consequently along the perpendicular bisector, away from the charges. It is zero only at the midpoint, not at every point on the bisector.

Correct answer: A. Along the perpendicular bisector towards the charges

Explanation: At a point on the perpendicular bisector, the two equal negative charges are at equal distances, so their field magnitudes are equal. The components parallel to the line joining the charges are opposite and cancel. Because the field due to a negative charge points toward that charge, the remaining perpendicular components both point toward the pair and add. Hence the net field is along the perpendicular bisector toward the charges. It is not always zero; only the midpoint has zero field by symmetry.

Correct answer: A. To define direction clearly and avoid disturbing the source arrangement

Explanation: Electric field is defined as E = F/q₀, where q₀ is a test charge. It is chosen positive so that the field direction is defined as the direction of force on a positive charge; a negative test charge would experience force in the opposite direction. It is chosen very small so that its own field does not significantly rearrange or disturb the source charges. These are separate requirements: positivity fixes direction, while smallness preserves the original field.

Correct answer: A. Magnitude remains same, direction reverses

Explanation: The electric field of a point charge is E = k|q|/r² for magnitude, so it depends on the magnitude of charge and the distance, not on whether the charge is positive or negative. Reversing the sign changes the field direction: a positive charge produces an outward field, while a negative charge produces an inward field. Since |q| and r are unchanged, the magnitude remains 20 N/C and only the direction reverses. Therefore, option A is correct; the other choices incorrectly change the magnitude or make it zero.

Correct answer: A. Like a single point charge with the total positive charge

Explanation: At distances much larger than the separation between the two charges, the detailed separation is not important in the leading approximation. The total charge is Q + Q = 2Q, which is non-zero and positive. Therefore, the dominant far-field term is the same as the field of a single point charge carrying total charge 2Q, with magnitude approximately k(2Q)/r². It is not a dipole field because the net charge does not vanish. Hence option A is correct.

Correct answer: A. A qualitative indication of larger charge or stronger field

Explanation: Electric field lines are an illustrative model, not physical threads that can be counted in space. Their density is used qualitatively: closely packed or more numerous lines in a region represent a stronger electric field, while the number of lines associated with a source can indicate a larger magnitude of charge. The drawing scale is arbitrary, so counting the displayed lines is not an exact measurement. Therefore option A is correct; the other choices confuse the representation with particles or material colour.

Correct answer: A. Direction may be the same, but magnitude is changing

Explanation: The tangent to an electric field line gives the field direction. Thus straight, parallel lines indicate that the direction can remain the same throughout the region. However, the spacing of field lines represents field magnitude qualitatively: unequal spacing means the line density, and therefore the field strength, changes from place to place. A uniform field requires both constant direction and constant magnitude, normally shown by equally spaced parallel lines. Hence option A is correct; options B, C, and D do not follow from the diagram.

Correct answer: A. Electric field direction at a point is unique

Explanation: At any particular point in space, the electric field vector has one definite direction. A field line is drawn so that its tangent gives this direction. If two electric field lines crossed, the intersection point would have two different tangents and therefore two different electric-field directions, which is impossible for a single-valued field. Lines can approach closely where the field varies rapidly, but they cannot intersect. Hence option A is correct; the other choices are unrelated or false.

Correct answer: A. In the same direction

Explanation: Electric fields obey vector addition. Since the two fields have equal magnitude and point in the same direction, the resultant field has magnitude 50 + 50 = 100 N/C and points in that same direction. The force on a charge is F = qE. For a positive test charge, q is positive, so the force direction is the same as the resultant electric field. A negative charge would experience the opposite direction, but that is not the case here. Therefore option A is correct.

Correct answer: A. Principle of superposition

Explanation: The principle of superposition states that the net electric field at a point is the vector sum of the fields produced there by all individual charges: E_net = E₁ + E₂ + ... . Two fields can each be non-zero yet cancel when they have equal magnitudes and opposite directions, giving a zero resultant. This is a vector cancellation, not a statement about conservation of charge, energy, or mass. Therefore option A is correct and directly explains the situation.

Correct answer: A. 0.0012 newton

Explanation: The governing relation is F = qE, because electric field is defined as force per unit positive test charge. Convert the charge first: 2 microcoulomb = 2 × 10^-6 coulomb. Therefore F = (2 × 10^-6)(600) = 1.2 × 10^-3 N = 0.0012 N. Since the charge is positive, the force is along the field direction. Thus option A is correct; B, C, and D result from incorrect unit conversion or calculation.

Correct answer: A. 2000 newton per coulomb

Explanation: Electric field is defined by E = F/q, so the force must be divided by the charge rather than multiplied by it. Convert 4 microcoulomb to 4 × 10^-6 C. Hence E = 0.008/(4 × 10^-6) = 2000 N/C. The positive sign of the charge affects force direction, but the requested field magnitude is positive. Therefore option A is correct; option B misses a factor of four, while C and D use an incorrect operation or scale.

Correct answer: A. Westward

Explanation: The governing vector relation is F = qE. For a positive charge, force and field point in the same direction, but for a negative charge they point in opposite directions because q is negative. The observed force is eastward, so the electric field must be westward. Its magnitude is not needed. Option A is therefore correct; option B incorrectly copies the force direction, and C and D have no basis in the stated information.