Electric Field and Electric Field Lines

At the midpoint, the field produced by the left positive charge points away from it, while the field from the right positive charge points in the opposite direction. Because the charges are equal and the distances are equal, the two field vectors have equal magnitudes and cancel, giving E_net = 0. Since line density represents field magnitude, high density there would be misleading.

The electric force on a charge placed at a point is F = qE, where E is the externally produced electric field at that point. If E = 0, then F = q × 0 = 0 for either a positive or a negative test charge, regardless of equal magnitude. The charge sign would reverse the force direction only when a nonzero field exists; it cannot create force in a zero field.

Electric field is a vector, so perpendicular fields must be combined using the Pythagorean relation rather than ordinary addition. If their magnitudes are E1 = 5 N/C and E2 = 12 N/C, then E = √(E1² + E2²) = √(25 + 144) = √169 = 13 N/C. Therefore option A is correct; 17 N/C is the simple arithmetic sum.

East and north are perpendicular directions, so the two field components form the legs of a right triangle. The resultant magnitude is E = √(E_east² + E_north²) = √(7² + 24²) = √(49 + 576) = √625 = 25 N/C. The resultant points northeast, but its magnitude is 25 N/C, making option A correct.

Let each field have magnitude E and let the angle between them be θ = 60°. The vector-addition formula gives R = √(E² + E² + 2E² cos θ). Since cos 60° = 1/2, R = √(2E² + E²) = √3E. Thus the resultant is √3 times either field. It is not 2E, which occurs only when the fields are parallel.

Let both fields have magnitude E and let the included angle be 120°. Their resultant is R = √(E² + E² + 2E² cos 120°). Because cos 120° = −1/2, R = √(2E² − E²) = √E² = E. Therefore the resultant has the same magnitude as either field. Zero would occur for equal fields at 180°, not at 120°.

The superposition principle states that the net electric field is the vector sum of the fields produced by all sources. Initially, equal fields in opposite directions cancel: E_net = E − E = 0. Removing one source removes one of these contributions, so the cancellation disappears and E_net equals the field due to the remaining source, with its original direction. Hence option A is correct.

The governing idea is Gauss’s law: the net electric flux through a closed surface is proportional to the net charge enclosed. Equal numbers of lines entering and leaving suggest zero net flux and therefore possibly zero net enclosed charge. However, flux is a surface integral, not the field at each point. An external charge can produce a nonzero field on the surface while its total flux is zero. Thus A is correct; B confuses zero net flux with zero field, and D is false.

By the superposition principle, the net electric field at a point is the vector sum of the fields produced by all charges. At a particular location, contributions from different sources can have equal magnitudes and opposite directions, making the resultant field zero. Nearby, the magnitudes or directions need not cancel, so the field can be nonzero. Thus A is correct. A single zero-field point does not prove the region is charge-free, and field lines never intersect in a regular field diagram.

The electric force on a charge is F = qE. For a positive test charge, the force has the same direction as the electric field, so right-pointing arrows produce a rightward force. Field-line density represents field magnitude; lines becoming denser toward the right indicate that E increases in that direction. Consequently, the force magnitude can increase as the charge moves right. A is correct; B reverses the direction, C assumes a uniform field, and D wrongly treats non-uniformity as zero field.

The electric force is F = qE. Because an electron has negative charge, its force is opposite to the direction of the electric field; upward field lines therefore produce a downward force. The lines become sparser upward, which indicates that the field magnitude decreases in that direction. Since the electron’s charge magnitude is constant, |F| = eE also decreases upward. Hence A is correct; B uses the positive-charge rule, C reverses the magnitude trend, and D is unjustified.

A uniform electric field requires both its direction and its magnitude to be constant throughout the region. Nearly parallel field lines indicate that the direction is nearly constant, but unequal spacing indicates that the field strength changes from place to place: closer lines represent larger magnitude and wider spacing smaller magnitude. Therefore the field is not uniform, so A is correct. B checks only direction, C is merely a general property of field lines, and D is false because electric field is a vector.

The governing relation is F = qE, so for the same positive charge q, the force magnitude is directly proportional to the electric-field magnitude. In a field-line diagram, greater line density represents a stronger field. Since the lines are closer near A, E is larger there and the charge experiences greater force at A. The charge sign fixes the direction, not this comparison of magnitudes; therefore B, C, and D are incorrect.

A field line is drawn so that its tangent at any point gives the direction of the electric field at that point. If the tangent direction changes from one nearby point to another, the field direction changes with position, which is why the line is curved. This does not imply zero magnitude or intersecting lines. Electric field is a vector, not a scalar, so A is the only valid conclusion.

At any point, the tangent to an electric field line specifies the direction of the electric field. For a positive test charge, F = qE, so its force has the same direction as E. If two field lines crossed, their two tangents would assign two different field and force directions to the same point. A single force cannot have two directions simultaneously; hence electrostatic field lines cannot intersect.

The electric force on a charge is F = qE. Because q is positive, the force on this charge points in the same direction as the uniform electric field, namely eastward. While the external force is present it balances the electric force, keeping the charge at rest. Immediately after removal, the eastward electric force is unbalanced, so Newton’s second law gives an eastward acceleration. Thus A is correct.

The electric force is given by the vector relation F = qE. An electron has negative charge, so its force is opposite to the direction of the electric field. The field here points westward; therefore the force on the electron points eastward. Since its mass is positive, Newton’s law a = F/m gives acceleration in the same direction as that force. Hence the electron accelerates eastward, not westward or with zero acceleration.

For a charge in an electric field, the force magnitude is |F| = |q|E. A proton carries charge +e and an electron carries charge −e, so their charge magnitudes are both e. In the same uniform field, each therefore experiences force magnitude eE. Their force directions are opposite because their charge signs differ, but direction does not alter the magnitude. Thus option A is correct; the forces are not zero unless the field is zero.

The density of field lines represents the strength of the electric field: closer lines indicate a larger E. However, the force on a particular charge is F = qE, so it depends on both the field and the charge magnitude. A very small charge can therefore experience a small force even in a strong-field region. Option A is correct. Dense lines do not mean that every charge feels the same force, and they certainly do not imply zero field or a positive charge.

The electric field direction is defined as the force direction on a positive test charge. Therefore a negative charge experiences electric force opposite to the field direction. Since the field points south, the electric force on the negative charge points north. For rest, the net force must be zero, so the external force must have equal magnitude and point south. Thus option A is correct; option B would add to the electric force, and the sideways or zero-force choices cannot produce equilibrium.

For a point charge, the field magnitude is E = k|Q|/r², so it varies inversely as the square of distance. Let the distances be rA and rB. Given EA = 9EB, we have (kQ/rA²)/(kQ/rB²) = (rB/rA)² = 9. Taking the positive square root because distances are positive gives rB/rA = 3, or rA = rB/3. Thus option A is correct; using 9 directly would ignore the square-root step.

The field of a point charge is E = kQ/r². If Q becomes 4Q, the charge change multiplies the field by 4. If the distance becomes 2r, the inverse-square factor changes it by 1/(2²) = 1/4. Combining both effects gives E' = k(4Q)/(2r)² = 4kQ/4r² = E. Therefore option A is correct. Considering only the charge or only the distance would give an incomplete result.

A positive charge experiences force in the direction of the electric field, so the upward-pointing lines give an upward force F = qE. Field-line crowding indicates field strength; lines becoming closer as height increases mean that E becomes larger upward. For the same positive charge, F = qE therefore also increases as it moves upward. Option A is correct. The force is not downward or zero, and it cannot remain constant when the field magnitude changes.

The field direction is defined independently of the particle’s instantaneous velocity. The electric force is F = qE: for a positive charge it points right, while for a negative charge it points left. A particle can nevertheless be moving left because velocity is determined by its prior motion and need not match the present force. Thus option A cannot be concluded from the information given.

Electric field lines are a visual representation, not physical strings or particle trajectories. Their arrows show the direction of the electric field, their relative density indicates comparative magnitude, and their starting or ending points reveal the signs and locations of charges. Reading all these features together is the most mature interpretation. Options B and D are irrelevant, while C confuses a visualization with actual motion.