Electric Field and Electric Field Lines

Correct answer: A. The field is very strong

Explanation: In an electric-field diagram, the number of lines crossing a given area represents the field-line density, which indicates the relative magnitude of the electric field. Very close spacing means many lines occupy a small area, so the field is strong there. It does not mean the field is zero; zero field would not be represented by a strong line density. The arrows still provide its direction.

Correct answer: A. Straight radial lines

Explanation: The governing concept is the field-line pattern of a point charge. A positive point charge produces an electric field directed radially outward, with spherical symmetry in three dimensions. Therefore, in a two-dimensional diagram, the lines appear as straight radial lines spreading away from the charge. Parallel lines represent a uniform field, while closed loops are not electrostatic field lines. Hence option A is correct.

Correct answer: A. To understand direction and relative strength of field

Explanation: Electric field lines are an imaginary visual representation, not physical threads. The tangent to a field line at any point gives the electric-field direction, and the density or closeness of lines indicates relative field strength. Thus they help us interpret both direction and comparative magnitude. They do not show colour, mass, or temperature. Therefore option A is the only scientifically appropriate answer.

Correct answer: A. Same

Explanation: In a uniform electric field, the magnitude and direction of the field remain constant from place to place. Field-line diagrams represent equal magnitude by drawing straight, parallel lines with equal spacing. Consequently, the line density is the same everywhere in the represented region. Increasing density would indicate a stronger field, whereas decreasing density would indicate a weaker field. Hence option A is correct.

Correct answer: A. By vector addition along the diagonal direction

Explanation: Electric field is a vector quantity, so perpendicular fields must be combined by vector addition rather than ordinary scalar addition. If the two components are represented by perpendicular sides of a right triangle or parallelogram, their resultant is along the diagonal. Its magnitude is E = √(E1² + E2²), and its direction depends on their ratio. Therefore option A is correct.

Correct answer: A. Because the two fields are in the same direction

Explanation: The electric field of a positive charge is directed away from that charge, whereas the field of a negative charge is directed toward it. At the midpoint, the distances from both charges are equal, so the two field magnitudes are equal. However, both vectors point from the positive charge toward the negative charge, so they add rather than cancel. Therefore option A is correct; equal magnitudes alone do not imply zero resultant.

Correct answer: C. Root two times one field

Explanation: For two perpendicular electric-field vectors, the resultant magnitude follows the Pythagorean relation: E_net = √(E₁² + E₂²). If both magnitudes are E, then E_net = √(E² + E²) = √2 E. It is therefore greater than one field but less than 2E. Option C is correct; simple addition applies only when vectors have the same direction.

Correct answer: A. There will be tangential field component and charges would move, so the diagram is wrong

Explanation: In electrostatic equilibrium, free charges in a conductor are at rest. If the electric field were oblique to the surface, it would have a tangential component, E_t, that exerts a force qE_t along the surface and makes charges move. Equilibrium therefore requires E_t = 0, so field lines must meet the conductor normally. Hence the shown oblique diagram is incorrect, making option A correct.

Correct answer: A. 3 to 1

Explanation: In a field-line diagram, greater line density represents greater electric-field magnitude, so the field ratio is approximately E₁:E₂ = 3:1. The force on a charge is F = qE. Because the same positive charge is placed at both points, q is common and cancels in the ratio: F₁:F₂ = E₁:E₂ = 3:1. Thus option A is correct.

Correct answer: A. The first point is at half the distance from the charge

Explanation: For a point charge, the electric-field magnitude is E = k|Q|/r², so it varies inversely as the square of distance. Let the second distance be r. If the first field is four times larger, kQ/r₁² = 4kQ/r², giving r₁² = r²/4 and r₁ = r/2. Therefore the first point is at half the distance, making option A correct.

Correct answer: A. Because direction changes with position

Explanation: A perfectly uniform electric field requires both constant magnitude and constant direction throughout the region. Equal spacing of field lines suggests that the magnitude may be approximately constant, but curved lines have tangents whose directions change from point to point. Therefore the field vector changes even if its magnitude is represented as constant. Option A is correct; equal spacing alone is insufficient for uniformity.

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

Explanation: Electric field-line direction gives the direction of the field, while the local density of lines represents its relative magnitude. Straight, parallel lines retain the same direction throughout the region. Because the spacing increases, fewer lines cross an equal area, so the field magnitude decreases in that direction. Therefore option A is correct. Option B would require uniform spacing; options C and D are not supported by the diagram.

Correct answer: A. North direction

Explanation: For a test charge q, the electric force is F = qE and acceleration is a = qE/m. Because the charge is positive, its force and acceleration point in the same direction as the electric field. Doubling the magnitude of q doubles the force and, for the same mass, the acceleration magnitude, but it does not rotate the field vector. Therefore the field still points north, so A is correct; B reverses the direction and C and D are not physical conclusions.

Correct answer: C. 10 newton per coulomb

Explanation: Electric field is a vector, so perpendicular contributions must be combined by the Pythagorean theorem rather than ordinary addition. If E1 = 8 N/C and E2 = 6 N/C, then E_net = √(E1² + E2²) = √(8² + 6²) = √(64 + 36) = √100 = 10 N/C. Therefore option C is correct. Option B would apply simple addition, A gives the difference, and D multiplies the magnitudes.

Correct answer: D. 9 to 1

Explanation: The electric field of a point charge is E = kQ/r², so its magnitude varies inversely as the square of distance. Let the first distance be r; the second is 3r. Then E1 = kQ/r², while E2 = kQ/(3r)² = kQ/(9r²) = E1/9. Hence E1:E2 = 9:1, making option D correct. Options A and C ignore the square dependence, while B reverses the required ratio.

Correct answer: A. No nearby shown negative charge is present to terminate the lines

Explanation: By convention, electric-field lines originate on positive charges and terminate on negative charges or extend to infinity. If the diagram shows lines leaving a positive charge and continuing outward without ending on a nearby negative charge, the displayed arrangement is consistent with a net positive source or an isolated positive charge in the shown region. Therefore A is the best inference. B is unsupported, C contradicts electrostatics, and D reverses the usual direction for lines from a positive charge.

Correct answer: A. Field of an isolated negative charge

Explanation: Electric-field lines point in the direction a positive test charge would move. They terminate on negative charges, so an isolated negative point charge has lines directed radially inward toward it; in a finite drawing those lines appear to arrive from infinity. This matches option A. An isolated positive charge has lines leaving it, a uniform field is represented by parallel equally spaced lines, and zero field has no meaningful field-line pattern. Thus the other options do not fit.

Correct answer: A. Draw the tangent to the line at that point

Explanation: A field line is a graphical curve whose tangent at any point gives the instantaneous direction of the electric field at that point. Near a charge, the curve can change direction rapidly, so the overall orientation or total length of the line is not sufficient. One should draw or visualize the tangent and follow its arrow. Therefore A is correct; color, line length, and naming the charge do not determine the local field direction.

Correct answer: B. No, because at the midpoint both fields are in the same direction

Explanation: The governing concept is superposition and the direction of the electric field. Between unlike charges, the field due to the positive charge points away from the positive charge, while the field due to the negative charge points toward the negative charge. At the midpoint these directions are the same, so the magnitudes add rather than cancel. Hence the net field cannot be zero there, regardless of the unequal magnitudes.

Correct answer: A. Because electrostatic field lines are linked with source and sink charges

Explanation: Electrostatic field lines originate on positive charges, where the field emerges, and terminate on negative charges, where the field enters; if no opposite charge is available, they may extend to infinity. Thus they have sources and sinks rather than forming closed loops. Equivalently, the electrostatic field is conservative and its circulation around a closed path is zero, unlike the directional pattern of magnetic field lines.

Correct answer: A. Force is to the right but its magnitude may decrease ahead

Explanation: Electric-field lines point in the direction of the electric field, so lines directed to the right imply a rightward field. A positive charge experiences force in the same direction because F = qE with q > 0. Decreasing line density indicates that the field magnitude becomes smaller toward the right. Therefore the force remains rightward while its magnitude may decrease as the charge moves ahead.

Correct answer: B. Leftward and increasing magnitude

Explanation: The field direction is left to right, so the electric field vector points rightward. The force on a charge is F = qE; for a negative q, the force points opposite to E, namely leftward. Increasing field-line density means that the field magnitude increases in that region. Since |F| = |q|E for a fixed charge, the force magnitude also increases. Hence option B is correct.

Correct answer: B. North direction

Explanation: The electric force is given by F = qE. For a negative charge, q is negative, so the force direction is opposite to the electric-field direction. If the field points south, the force on the released negative charge points north. Since acceleration has the direction of the net force through a = F/m, the initial acceleration is northward; therefore option B is correct. A would apply to a positive charge, while C and D have no basis.

Correct answer: A. Magnitude doubles and direction is opposite to field

Explanation: For a charge in an electric field, the force magnitude is |F| = |q|E. The charge is unchanged, so doubling E doubles the force magnitude. The force direction is determined by the sign of q; because the charge is negative, it is opposite to the field direction both before and after the change. Thus option A is correct. The other options incorrectly predict halving, no change, or zero force.

Correct answer: A. Because they are imaginary lines showing field direction and density

Explanation: Electric-field lines are an imaginary graphical representation: their tangent gives the field direction, and their relative density indicates field strength. They are not material tracks followed by charged particles. A particle’s actual trajectory depends on its initial velocity, charge, mass, and the net force, and it may cross or deviate from a field line. Hence option A is correct; B and C misidentify their meaning, while D denies the real effect of the field.