Electric Field and Electric Field Lines

Electric field is a vector, so opposite components must be added algebraically with opposite signs. Take east as positive: E_net = +10 N/C − 6 N/C = +4 N/C. The positive sign means the resultant points eastward, in the direction of the larger field. Therefore A is correct. Adding the magnitudes would be appropriate only for fields in the same direction; zero would require equal opposite magnitudes.

Use vector superposition by resolving the fields into perpendicular components. The eastward component has magnitude 2 and the westward component also has magnitude 2, so they cancel: E_x = 2 − 2 = 0. The remaining northward component is 5, giving a nonzero resultant directed north. Hence A is correct. It is not zero because the northward field has no equal southward counterpart.

By definition, the electric field at a point has the direction of the force exerted on a positive test charge placed there. Therefore, if E points northeast, the force F = qE on a positive charge also points northeast. The force is not reversed because q is positive. Southwest would apply to a negative charge, while north or east alone would ignore one component of the stated direction.

Electric field lines originate on positive charges and terminate on negative charges. Consequently, lines directed outward from a charge identify it as positive, while lines directed inward toward a charge identify it as negative. The first charge therefore has negative sign and the second has positive sign. Options C and D incorrectly assign the same sign to both charges, while option B reverses the two identifications.

A uniform electric field requires both its magnitude and its direction to remain constant from point to point. In this case, the magnitude is unchanged, but the direction varies, so the field is non-uniform. Therefore, option A is correct. Option B is wrong because equal magnitude alone is insufficient; option C is unrelated, since curved lines generally indicate changing direction; and option D is wrong because electric field is a vector quantity.

Electric field is a vector, so perpendicular fields cannot be added by ordinary arithmetic. They form the two perpendicular sides of a right triangle. Thus E = sqrt(E1² + E2²) = sqrt(3² + 4²) = sqrt(9 + 16) = sqrt25 = 5 N/C. Therefore option A is correct. Option B is the direct sum, while C is the difference and D has no valid vector basis.

Oppositely directed electric fields are represented by vectors on one straight line, so their magnitudes are subtracted. Taking east as positive, E_net = +8 - 3 = +5 N/C. The positive sign means the result points eastward; hence option A is correct. Option B incorrectly adds opposite vectors, option C reverses the direction, and D makes both errors.

The electric field at a point is defined as the force per unit positive test charge, E = F/q₀. Since q₀ is positive, dividing the force vector by q₀ does not reverse its direction. Therefore, if the force on the positive test charge is westward, the electric field is also westward. The eastward answer would apply to a negative charge experiencing a westward force, not to the positive test charge specified here.

The force on a charge is related to the electric field by F = qE. For a negative charge, q is less than zero, so the force is opposite to the field direction. The given force points westward; therefore the electric field must point in the opposite direction, eastward. Westward would incorrectly assume a positive charge, while northward and southward have no basis in the stated force direction.

Electric field lines are an illustrative representation, and their density indicates the relative magnitude of the field. Where the lines are closer together, more lines cross a given area, representing a stronger electric field. Therefore, if the spacing gradually decreases along a direction, the field magnitude is increasing in that direction and the field is non-uniform. Equal spacing would indicate a uniform field; closer spacing does not mean the field is becoming zero.

By convention, electric field lines originate on positive charges and terminate on negative charges. Near the starting source, lines directed outward indicate that a positive charge is producing the field. Their later bending toward a negative charge is consistent with the field configuration between opposite charges and does not change the identity of the source. A negative source would have lines directed inward near itself. Therefore the starting charge is positive, so option A is correct; neutrality cannot produce outward-originating lines.

The governing relation is F = qE. Dense field lines indicate that the electric field magnitude E is large, but an electric force still requires electric charge q. For an uncharged particle, q = 0, so F = 0 multiplied by E = 0, even when E is strong. Options B, C, and D would be relevant only after specifying a nonzero charge and its sign.

Electric field lines are drawn in the direction of the force on a hypothetical positive test charge. They originate at positive charges and terminate on negative charges, or extend to infinity when no negative charge receives them. Thus a line beginning at one positive charge cannot correctly end at another positive charge. Positive charges also repel, so option C reverses the interaction.

The governing definition is E = F/q, where F is the force on a small positive test charge q. If q is doubled and the measured force F also doubles, their ratio remains unchanged: E' = 2F/2q = F/q = E. Thus the test charge changes the force proportionally, not the source-produced field, provided it is sufficiently small and does not disturb the sources. Therefore option A is correct; B confuses force with field, while C and D do not follow from the observation.

Electric field at a point is defined as the force per unit positive test charge, with the test charge assumed small enough not to alter the original charge distribution. A large test charge can exert its own electric force on nearby source charges, shifting them or changing their arrangement. The measured force would then correspond to a changed configuration rather than the original field. Hence option A is correct. A test charge does not make the field scalar, force zero, or distance infinite.

For a positive charge, the electric force is in the same direction as the electric field, so an upward field produces an upward force F = qE. To keep the charge at rest, the net force must be zero. The external force must therefore have the same magnitude as qE but point downward, directly opposing the electric force. Option A is correct. An upward, rightward, or leftward force cannot cancel the upward electric force and would not give equilibrium.

The governing rule is that electric-field lines originate at positive charges and terminate on negative charges, or continue to infinity if no negative charge is available. Therefore, the line may correctly start at the first positive charge, but it cannot end at the second positive charge. Option B is false because every charge creates an electric field, option C reverses the rule, and option D contradicts like-charge repulsion.

Electric-field direction is defined as the force direction on a small positive test charge. Consequently, field lines leave positive charges and enter negative charges; a negative charge acts as a sink. If arrows are drawn outward from a negative charge, the direction rule is reversed, so option A identifies the violation. Lines may be curved, they cannot intersect at one point, and their density does indicate relative field strength.

Electric field is a vector quantity, so both its magnitude and its direction matter. A uniform electric field has the same vector value at every point: its magnitude remains constant and its direction also remains unchanged. If the magnitude is constant but the direction varies with position, the vector changes and the field is non-uniform. Therefore option A is correct; the other options ignore the vector nature of E.

The electric force is given by F = qE. With the field fixed, the force magnitude is proportional to the magnitude |q|, so doubling |q| doubles |F|. The direction depends on the sign of the charge: a negative charge experiences force opposite to the field. Since the field points northward, the force points southward. Therefore option A is correct.

The governing relation is F = qE. Using the magnitude of charge, |F| = |q|E = (3 × 10^-6 C)(2000 N/C) = 6 × 10^-3 N = 0.006 N. Because the charge is negative, its force is opposite to the electric field direction. The field points eastward, so the force points westward. Option B has the right magnitude but the wrong direction, while C and D use incorrect arithmetic or units.

The arrow on an electric field line represents the direction of the electric field at that point. The electric force is F = qE, and for a positive charge q is positive, so the force is in the same direction as E. Since acceleration has the direction of the net force, the initially released positive charge accelerates eastward. A westward answer would apply to a negative charge, not a positive one.

The field-line arrow gives the direction of the electric field, which is eastward here. The force on a charge is F = qE. Since q is negative, the force reverses direction and points westward. The initial acceleration follows the force because a = F/m and mass is positive. Thus option A is correct. Eastward would be correct for a positive charge, whereas zero acceleration is not implied by merely releasing the charge.

The direction of a field line represents the direction of the electric field, while its spacing or density is used to represent field magnitude. In a uniform electric field, both magnitude and direction remain constant from point to point. Therefore, the lines are parallel and equally spaced: they neither spread apart nor converge. Uniformity does not mean the field has no direction or that charges cannot exist.

The electric force on a charge is given by F = qE. An electron has q = −e, so its force vector is opposite to the electric-field vector. If E points eastward, the force on the electron points westward; hence option A is correct. A positive charge would be pushed eastward, but the force is not zero merely because the particle is an electron.