Electric Field and Electric Field Lines

Use F = qE. At the same place the field is unchanged, and equal-magnitude charges produce forces with equal magnitudes because |F| = |q|E. Changing the charge from positive to negative reverses the direction. Thus a negative charge experiences a force of the same magnitude southward. Option A is correct; B ignores the sign change, C incorrectly doubles the force, and D wrongly assumes cancellation without another force.

Electric-field-line arrows indicate the direction of the electric field, defined as the force direction on a positive test charge. In a uniform region, parallel lines should have the same direction and approximately equal spacing. Opposite arrows in the same nearby region assign two directions to one field, so the representation is inconsistent. Option A is correct; parallel lines can represent a uniform field, arrows are meaningful, and electrostatic fields are not always circular.

Electric field lines are an illustrative representation, not physical wires or a complete set of lines filling space. Their number and spacing are chosen to show direction and relative strength. A diagram may omit lines for clarity, so an unmarked region does not automatically have zero field. Option A is correct. Option B treats representative lines as real boundaries, C makes the same misconception, and D incorrectly denies electric fields.

The electric field of a positive charge points away from it, while the field of a negative charge points toward it. At every point between unlike charges, both fields therefore point from the positive charge toward the negative charge. They add rather than cancel, regardless of whether their magnitudes are equal. Hence the midpoint cannot be a zero-field point; option A is correct. Option D is wrong because the midpoint is not at zero distance from either charge.

For a point charge, the electric-field magnitude is E = k|Q|/r². Replacing Q by 3Q multiplies the field by 3. Replacing r by r/2 multiplies the inverse-square factor by 1/(1/2)² = 4. Therefore E' = k(3Q)/(r/2)² = 3 × 4E = 12E. The correct answer is option A; option C ignores the distance change, while option D ignores the charge change.

The field of a point charge follows E = k|Q|/r². Increasing the source charge from Q to 9Q multiplies E by 9. Increasing the distance from r to 3r changes the inverse-square factor by 1/3² = 1/9. Combining both changes gives E' = (9/9)E = E. Thus the field remains unchanged, so option A is correct. Options B and C consider only the charge increase, whereas option D uses an incorrect distance dependence.

At any point, the tangent to an electric field line gives the direction of the electric field there. If two field lines intersected, their tangents at the intersection would assign two different directions to the same point. A physical electric field has one unique resultant direction at a given point, except at a zero-field point where direction is undefined rather than two-valued. Therefore field lines cannot intersect, and option A identifies the error.

For a point charge, Coulomb’s law gives E = (1/4πε₀)|q|/r². The field lines spread over spherical surfaces whose area is 4πr², so the same total flux is distributed over an area that grows as r². Hence field-line density and electric-field magnitude decrease as 1/r². Option A expresses this inverse-square law; the other choices have the wrong dependence.

By definition, the tangent to an electric field line gives the direction of E at that point. The electric force is F = qE. For a negative charge, q < 0, so the force vector and therefore the acceleration point opposite to E, assuming the mass is positive. The curvature of the line does not change this local sign rule. Hence A is correct; a negative charge can experience force, and circular motion is not automatic.

Electric field is a vector, so uniformity requires both its magnitude and its direction to remain unchanged throughout the region. Here the magnitude is constant, but the direction continuously rotates from point to point. Therefore the vector field is non-uniform, even though its strength is unchanged. Option B is incorrect because constant magnitude alone is insufficient; option C says the magnitude is zero, which is not given, and option D is not a valid physical classification.

Let the two perpendicular fields have equal magnitude E and act along the x- and y-axes. Their resultant has components E and E, so its direction satisfies tan theta = E/E = 1. Hence theta = 45 degrees measured from either axis. Equal components make the resultant lie exactly halfway between the two fields. Thirty or sixty degrees would require unequal components, while ninety degrees would mean the resultant followed only one field.

At the centre of an equilateral triangle, the distances from all three vertices are equal. Consequently, the three positive charges produce electric fields of equal magnitude. Their directions are separated symmetrically by 120 degrees, and the vector sum of three equal vectors arranged this way is zero. Thus the net electric field is zero. A direction toward a corner would break the threefold symmetry, so options B, C, and D are not justified.

The centre of the square is at the same distance from all four vertices, so equal charges produce equal field magnitudes there. Consider each pair of opposite corners: the fields from that pair have equal magnitudes but opposite directions, so they cancel. The other opposite pair cancels in the same way, leaving zero net field. Positive charges do produce fields, the centre is not at zero distance, and square geometry does not prevent field lines.

The electric field points in the direction of the steepest decrease of electric potential, expressed by E = -∇V. Along a specified line, if the potential decreases continuously in one direction, the field component along that direction is positive and points generally toward lower potential. A positive test charge is therefore pushed from higher potential toward lower potential. Option B reverses the sign, option C is not generally true, and option D would apply only if the potential had no spatial change.

The governing relation is F = qE. An electron has negative charge, so its electric force is opposite to the electric-field vector. Because it is released from rest, its initial acceleration is in the direction of that force, namely opposite to the field lines. It does not move along the lines as a positive charge would, and a uniform field still exerts a nonzero force. Thus option A is correct.

For a positive charge, F = qE and Newton’s law gives a = qE/m. Since q and m are positive, acceleration has the same direction as the electric field. Halving q changes the force and acceleration magnitudes, but it does not reverse their direction or make the external field zero. Therefore an eastward acceleration indicates an eastward electric field, so option A is correct.

The field direction is defined as the force direction on a positive test charge. The given negative charge feels a northward force, so the electric field at that point must point southward. Replacing it with a positive charge of equal magnitude makes the force parallel to the field, hence southward. With equal absolute charge, the force magnitude is unchanged; only its direction reverses. Option A is correct.

In a field-line diagram, greater line density conventionally represents greater electric-field magnitude. The force on a charge is F = qE. For the same positive q, increasing E increases the force magnitude in direct proportion. Therefore denser lines should normally imply a larger, not smaller, force on that charge. Option A states this relation; the other options incorrectly reject the meaning of line density or electric force.

Electric field lines are a visual representation, not physical objects whose absolute count is fixed by nature. Their number is chosen according to a drawing scale, although relative density and direction convey field information. If the same point charge is represented with twice as many lines, the scale or convention has changed; it does not prove that the charge doubled. Hence option A is correct, while B, C and D do not follow.

The direction convention for electrostatic field lines is from positive charge toward negative charge. A positive charge acts as a source, so lines emerge from it; a negative charge acts as a sink, so lines terminate on it. Therefore lines shown emerging from the negative charge must be reversed so that they enter that charge. Option A is correct. Closed circles describe neither ordinary electrostatic field lines nor this source-sink behavior.

The two field vectors are perpendicular because east and north differ by 90 degrees. Therefore the resultant magnitude follows the Pythagorean theorem: E = √(E₁² + E₂²) = √(5² + 12²) = √(25 + 144) = √169 = 13 N/C. Adding 5 and 12 directly would incorrectly treat the vectors as parallel. Hence option A is correct.

The governing equation is F = qE. For a negative charge, q is negative, so the force direction is opposite to the electric-field direction. The given force points southwest; reversing that vector gives northeast. The force magnitude of 13 N does not affect the required direction, and the field is not in the same southwest direction because that would apply to a positive charge. Thus option A is correct.

The governing idea is that the density of electric field lines represents the magnitude of the electric field, not a separate physical force. For a charge q, the electric force is F = qE. Because the same positive charge enters a region where E is larger, the magnitude of its force can increase. Its direction follows the local field direction, but it need not reverse; its charge also cannot disappear merely because line density changes.

For a point charge, the electric force is given by F = qE. Since q is positive, the force has the same direction as the electric field; therefore it points eastward. The field is non-uniform and becomes stronger farther east, so the magnitude qE increases as the charge moves ahead, assuming the charge remains the same. It is not zero and is not perpendicular to E.

Electric field is a vector quantity, so it is specified by both magnitude and direction. A uniform electric field must have the same vector value at every point: neither its strength nor its direction may vary spatially. A large magnitude alone is insufficient. Repeated changes in direction make the field non-uniform even if the magnitude happens to remain constant.