Electric Field and Electric Field Lines

Correct answer: A. Stronger

Explanation: The governing diagrammatic convention is that the density of electric field lines represents the magnitude of the field. When lines are closer together, more lines pass through the same area, indicating a larger electric field and hence a greater force per unit positive charge. Widely spaced lines indicate a weaker field. Thus option A is correct, while zero or directionless fields are not implied by close spacing.

Correct answer: A. Equally spaced parallel straight lines

Explanation: A uniform electric field has the same magnitude and the same direction at every point in the region. Field lines therefore must be straight and parallel to show a constant direction, and equally spaced to show a constant magnitude. The nearly uniform field between large, oppositely charged parallel plates is a familiar example. Curved, circular, or unevenly spaced lines represent non-uniform situations, so option A is correct.

Correct answer: A. Remains unchanged

Explanation: The governing relation is E = F/q, where E is the field at the location, F is the force on the test charge, and q is that charge. Initially E = F/q. After both quantities double, E' = 2F/2q = F/q = E. Thus the electric field remains unchanged, provided the test charge is sufficiently small and does not disturb the source-charge arrangement. Option A is correct; changing only force or only charge would give a different result.

Correct answer: A. Becomes one-fourth

Explanation: For a point charge, the electric-field magnitude is E = k|Q|/r². If the distance changes from r to 2r while Q remains fixed, the new value is E' = k|Q|/(2r)² = k|Q|/(4r²) = E/4. Hence the field becomes one-fourth of its original value. The half and double choices ignore the inverse-square dependence, while four times reverses the effect, so option A is correct.

Correct answer: A. Perpendicularly

Explanation: In electrostatic equilibrium, the electric field has no tangential component at a conductor’s surface. If a tangential component existed, free charges would experience a force along the surface and continue moving, contradicting equilibrium. The remaining field is normal to the surface, so electric field lines meet it perpendicularly. Parallel or arbitrary-angle lines would imply a tangential component, and curvature alone is not the governing condition. Thus option A is correct.

Correct answer: A. Because electrostatic field is conservative

Explanation: Electrostatic field lines do not form closed loops because an electrostatic field is conservative and its circulation around any closed path is zero. The lines originate at positive charges and terminate at negative charges, or extend to infinity, rather than returning to their starting point. Option A is therefore correct. The other options confuse electric field properties with mass, magnetism, or the incorrect claim that the field is always zero.

Correct answer: A. Zero

Explanation: At the midpoint, the distances from the two equal positive charges are identical, so each charge produces an electric field of the same magnitude, E = kq/r². The field due to the left charge points away from it, while the field due to the right charge points in the opposite direction. Their vector sum is therefore zero, making option A correct. The fields do not add because their directions oppose each other.

Correct answer: A. Zero

Explanation: An electric field produced by a negative charge is directed toward that charge. At the midpoint, the two equal negative charges are at equal distances, so their fields have equal magnitudes, E = k|q|/r². One field points toward the left charge and the other toward the right charge; the directions are opposite. They cancel by superposition, so option A, zero, is correct. The field is not infinite because the midpoint is not on either charge.

Correct answer: A. From the positive charge to the negative charge

Explanation: Let the positive charge be on the left and the negative charge on the right. At the midpoint, the field from the positive charge points away from it, toward the right. The field from the negative charge points toward it, also toward the right. Thus both equal contributions point from positive to negative and add, rather than cancel. Option A is correct; zero field would apply to two equal like charges at their midpoint.

Correct answer: A. The direction of electric field at that point

Explanation: An electric field line is drawn so that its local direction represents the direction of the electric field, defined as the force direction on a positive test charge. When the line is curved, its tangent at a selected point gives the instantaneous field direction there. The tangent does not directly give charge, potential, or mass; field-line density may indicate relative field strength, but not the listed quantities. Therefore option A is correct.

Correct answer: A. So it does not disturb the source-charge arrangement

Explanation: A test charge is introduced to measure or define the electric field produced by source charges. It must be sufficiently small so that its own electric field and the force it exerts do not significantly rearrange or polarize the source charges. Then the measured force per unit charge represents the original field, E = F/q. Option A is correct. Its smallness is not intended to increase mass, remove the field, or make field lines closed.

Correct answer: A. Magnitude of the test charge placed there

Explanation: Electric field is defined as force per unit positive test charge, E = F/q. The field at a point is produced by the source charges and depends on their values, positions, the observation point, and the surrounding medium. If the test charge is changed, the force changes proportionally, but F/q remains the same, provided the test charge is sufficiently small and does not disturb the arrangement. Therefore, option A is correct; the other factors can affect the field.

Correct answer: A. Magnitude of charge

Explanation: Electric field lines are an illustrative model rather than individually countable physical objects. In a field-line diagram, the number of lines associated with a charge is chosen in proportion to the magnitude of that charge. More lines indicate a larger charge magnitude and, correspondingly, a stronger field near the charge; fewer lines indicate a smaller magnitude. This is qualitative, not an exact numerical measurement. Therefore option A is correct.

Correct answer: A. They bend away from each other

Explanation: Each positive charge produces field lines directed outward. For two equal positive charges, the fields oppose one another along the central line, and the surrounding field pattern bends outward away from the region between the charges. The lines do not join the two positive charges, because field lines do not terminate on another positive charge. They also never cross, since one point cannot have two field directions. Therefore option A is correct.

Correct answer: A. When the placed charge is positive

Explanation: The force on a charge is described by the vector equation F = qE. If q is positive, the scalar multiplication preserves the direction, so F and E point the same way. If q is negative, the force reverses direction. A zero charge experiences zero force, for which a direction is not defined, and field uniformity is unrelated to the sign relation. Therefore option A gives the required condition.

Correct answer: A. By vector addition of the two fields

Explanation: The principle of superposition states that the net electric field is the vector sum of the fields produced independently by all source charges: E_net = E1 + E2. Both magnitude and direction must be included. If the two fields are parallel and oppositely directed, their magnitudes may effectively subtract, but that is a special result of vector addition, not a universal rule. Ignoring one charge is incorrect. Therefore option A is correct.

Correct answer: A. High

Explanation: The density of electric field lines is a visual representation of electric-field strength. For a point charge, the magnitude is E = k|q|/r², so it becomes very large when the distance r is small. Therefore, lines are drawn closer together near the charge, indicating a strong field. They are not zero or uniformly spaced at every distance; their spacing increases as the distance grows.

Correct answer: A. It decreases

Explanation: For a point charge, the electric-field magnitude follows the inverse-square relation E = k|q|/r². As the distance r from the charge increases, the field becomes weaker. Field-line diagrams represent a weaker field by drawing the lines farther apart, so their apparent density decreases with distance. It does not suddenly become zero, and it is not constant at all distances.

Correct answer: A. Direction of electric field

Explanation: An electric-field line is defined so that its tangent and arrow indicate the direction of the electric field at each point. Equivalently, the arrow gives the direction of force on a small positive test charge placed there, because F = qE for positive q. It does not represent the charge’s mass, the drawn line’s length, or the medium’s temperature.

Correct answer: A. Field magnitude is changing with position

Explanation: The convention for field-line diagrams is that line density represents the magnitude of the electric field, while the arrow or tangent represents its direction. If the spacing changes from one region to another, the density changes and therefore the field magnitude changes with position. This describes a non-uniform field. Variable spacing does not prove that the field is zero or that no charges exist.

Correct answer: A. Double and in the same direction

Explanation: Electric field is a vector, so the resultant must be found by vector addition. Let each field be E in the same direction. Then E_net = E + E = 2E, and the direction remains unchanged. Cancellation would occur only for equal fields in opposite directions. The result is not half or infinite under the stated conditions, so option A is correct.

Correct answer: A. Zero

Explanation: Electric fields combine by vector addition. If one field is +E and the other is −E, their resultant is E_net = E + (−E) = 0. Thus equal magnitudes in exactly opposite directions cancel completely. A doubled, halved, or four-times result would require different magnitudes or a different angular relationship; none applies here. Therefore, option A is correct.

Correct answer: A. Along the line joining the charge and that point

Explanation: The electric field produced by an isolated point charge is radial. Consequently, at any selected point, the field lies along the straight line joining the charge and that point. For a positive charge it points away from the charge; for a negative charge it points toward the charge. It is not always horizontal or vertical, and it is not tangential to a circle centered on the charge.

Correct answer: A. Along the field direction

Explanation: The electric force on a charge is F = qE. For a positive charge, q is positive, so the force has the same direction as the electric field. If the charge is released freely, this force produces acceleration in that direction, assuming no other force changes the motion. A negative charge would accelerate oppositely, but that is not the case here. Hence option A is correct.

Correct answer: A. Opposite to the field

Explanation: The electric force is given by F = qE. For a negative charge, q is negative, so the force vector points opposite to the electric-field vector. When released, the charge initially accelerates and begins moving in the force direction, provided no other force dominates. It therefore moves opposite to the field. It does not move without force or necessarily follow a circular path.