Electric Field and Electric Field Lines

Electric field lines are an illustrative tool: the tangent to a line gives the electric-field direction at that point. If two lines crossed, the same point would have two different directions, which is impossible for a uniquely defined vector field. At the midpoint, equal and opposite contributions cancel so the net field is zero; this does not create two valid directions or justify an intersection.

In electrostatics, field lines originate on positive charges and terminate on negative charges or extend to infinity. Equivalently, the electrostatic field is conservative, so its line integral around a closed path is zero and a field line cannot continuously circulate with a nonzero tangential field. Therefore option A gives the correct physical picture. Options B, C, and D are false: positive charges do produce fields, lines are not confined inside conductors, and the field is not zero everywhere.

By superposition, the net electric field is the vector sum of fields produced by all source charges. At a particular point, these vectors can have equal magnitudes and opposite directions, giving zero resultant. Moving slightly changes distances and directions, so the cancellation generally disappears and the field becomes nonzero. Thus option A is correct. Option B wrongly extends a local cancellation everywhere, while C and D do not follow from the observation.

For unlike charges, the electric fields between them point in the same direction, so cancellation is impossible in the inner region. A zero point can occur outside, where the two fields oppose one another. The cancellation point lies on the side of the smaller charge because it must be closer to that charge to compensate for the larger charge’s stronger field. Thus option A is correct. The midpoint cannot work because the fields there reinforce each other.

Let each field have magnitude E. The vector-addition formula is R = √(E² + E² + 2E² cos θ). For θ = 120°, cos 120° = −1/2, so R = √(2E² − E²) = √E² = E. Therefore the resultant has the same magnitude as either individual field, making option A correct. It is not 2E because the fields are not parallel, and it is not zero because they are not exactly opposite.

Two equal vectors cancel exactly only when their directions differ by precisely 180°. If one vector is rotated slightly, the vectors are no longer exactly opposite, so their sum acquires a nonzero component. The resultant may be small when the rotation is small, but it is not zero; its direction is also determined by the vector geometry. Hence option A is correct. Option B incorrectly assumes that equal magnitudes alone guarantee cancellation.

A perfectly uniform electric field must have both constant magnitude and constant direction throughout the region. Equal spacing of field lines can represent a constant magnitude, but the tangent to a curved field line changes from point to point. Since the tangent gives the local field direction, circular arcs indicate a changing direction and therefore a non-uniform field. Option A is correct. Curved field lines can exist in non-uniform fields, so option D is too absolute.

An equipotential surface has the same potential at every point. The relation dV = −E·dl shows that a component of electric field along a small tangential displacement would produce a nonzero potential change along the surface. Since dV must be zero for motion within an equipotential surface, the tangential component must vanish. The electric field is therefore normal to the surface. Option A is correct; the other statements confuse potential, charge, or vector properties.

A conductor contains mobile free charges. If an electric field had a tangential component at its surface, that component would exert a force qE_parallel on the free charges and drive them along the surface. Their redistribution would continue until the electrostatic equilibrium condition E_parallel = 0 was restored. Thus a nonzero tangential component cannot persist in electrostatics, and option A is correct. It does not turn the conductor into an insulator or remove all charge.

An electric field line is a graphical curve whose tangent gives the electric-field direction at each point; it is not automatically the trajectory of a particle. The motion follows Newton’s law, F = qE, and depends on the charge, mass, initial velocity, and how the field changes in space. A charge released from rest may initially move along the local tangent, but it need not remain on the whole curved line. Thus A is correct.

Imagine placing an equal positive charge at the empty corner as well. With four equal charges, symmetry would make the field at the centre zero. Therefore, the field produced by the remaining three charges must be equal and opposite to the field that the missing charge would have produced. A positive charge at the empty corner would produce a field at the centre directed away from that corner, so the remaining-field resultant points toward the empty corner.

Electric field and electric potential are different physical quantities. Their relation is E = -dV/dr in one dimension, or E = -∇V generally. Thus the field depends on the spatial rate of change of potential, not directly on its absolute value. At a point where the potential has a local maximum, minimum, or flat slope, E can be zero while V remains non-zero. Hence option A is correct; the other statements incorrectly equate the quantities or impose unsupported charge restrictions.

At the exact midpoint, the repulsive forces from the two equal positive charges cancel. Move the positive test charge slightly to the right: it is now closer to the right-hand source charge, so the repulsive force from that charge is larger because F = k|qQ|/r². That stronger force pushes the test charge leftward. The left charge also repels it rightward, but more weakly because it is farther away. Therefore the net force is to the left.

The governing concept is vector addition of electric fields. At a point above the midpoint, the field due to the positive charge points away from that charge, while the field due to the negative charge points toward it. The vertical components may cancel by symmetry, but both horizontal components point from the positive charge toward the negative charge and therefore add. Hence option A is correct; option D incorrectly assumes complete cancellation.

The governing relation is E = −∇V; in one dimension, E_x = −dV/dx. Thus the electric field points toward decreasing potential. If field lines point right, potential must decrease as position increases to the right. Claiming that potential increases from left to right gives the opposite sign and is inconsistent, so option A is correct. The other choices deny or reverse this fundamental relation.

A reliable field-line reading combines several pieces of information. Arrowheads give the local direction of E, and the relative density of lines indicates the field magnitude. For systems with several charges, symmetry helps identify cancellation or reinforcement, while vector superposition gives the resultant field. Line length is not a physical magnitude, and field lines are representations, not particle trajectories.

For two like charges, the electric fields between them point in opposite directions, so cancellation is possible only in the region between the charges. The larger charge produces the stronger field at equal distances; therefore, to make the two magnitudes equal, the zero-field point must be nearer the smaller charge, where its field is stronger relative to the distance. The midpoint works only for equal charges.

Electrostatic field lines represent a conservative electric field. They originate on positive charges and terminate on negative charges or at infinity; they do not circulate continuously back to their starting points. Equivalently, the line integral around a closed path is zero, and electrostatic potential is single-valued. Closed field-line patterns are associated with magnetic fields, not ordinary electrostatic fields, so A expresses the relevant idea.

The force magnitudes are equal because both particles have the same charge magnitude e, giving |F| = eE in the same field. Acceleration, however, follows Newton’s second law, a = F/m. A proton is far more massive than an electron, so the same force produces a much smaller proton acceleration and a much larger electron acceleration. Their acceleration directions are opposite, but the essential reason for unequal magnitudes is their different masses.

Electric field is a vector, whereas electric potential is a scalar. At the midpoint of two equal positive charges, the fields due to the charges have equal magnitudes and opposite directions, so their vector sum is zero. The potentials have equal positive values and add algebraically, giving a positive total potential, V = kQ/r + kQ/r. This distinction makes A correct; opposite charges would instead give zero potential at their midpoint.

Electric potential from equal opposite charges is V = kQ/r + k(−Q)/r, so the scalar contributions cancel at the midpoint. Electric field contributions must be added as vectors. At a point between the charges, the field from the positive charge points away from it, while the field from the negative charge points toward it; both point in the same direction, from positive to negative. They therefore reinforce rather than cancel, so A is correct.

Consider the two equal positive charges symmetrically placed on a line and a point on their perpendicular bisector. The field from each positive charge points away from that charge. Because the distances and charge magnitudes are equal, the components parallel to the joining line are equal and opposite, so they cancel. The perpendicular components point in the same direction and add, away from the charge line. Therefore option A is correct; the field is not generally zero.

Electric field lines terminate on negative charges, so the field produced by each negative charge points toward it. At a point on the perpendicular bisector, symmetry makes the components parallel to the charge-joining line equal and opposite; they cancel. The perpendicular components point toward the line containing the negative charges and reinforce one another. Hence option A is correct. The direction is opposite to the corresponding result for two equal positive charges, and the net field is not generally zero.

Electric field is a vector, whereas electric potential is a scalar. At the midpoint between two equal positive charges, the fields have equal magnitudes and opposite directions, so their vector sum is zero. Their potentials have the same positive sign and therefore add, giving a nonzero potential, specifically V = 2kq/r when each charge is at distance r. At infinity the potential is normally zero, and the other choices do not provide the stated condition.

For equal opposite charges, the scalar potentials at the midpoint are equal in magnitude and opposite in sign, so they cancel and the net potential can be zero. The electric field contributions behave differently because field is a vector. At the midpoint, the field due to the positive charge points away from it, and the field due to the negative charge points toward it; both point from the positive charge toward the negative charge. Hence the fields add rather than cancel.