Electric Field and Electric Field Lines

Correct answer: C. Taking the vector sum of both fields

Explanation: Electric field is a vector quantity, so both its magnitude and direction must be considered. For two charges, determine each field at the point and add them using vector addition; components may be resolved along perpendicular axes when necessary. Ordinary scalar addition is valid only when directions are appropriately accounted for. Therefore the net field is the vector sum, making option C correct; the other choices wrongly discard real contributions.

Correct answer: A. Zero

Explanation: Let the equal negative charges be placed symmetrically on a line, with midpoint P between them. At P, the field due to the left charge points toward that charge, while the field due to the right charge points toward the right charge. Their distances and magnitudes are equal, so the two opposite vectors cancel: E_net = E - E = 0. Hence option A is correct; the field is not directed toward one side or upward.

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

Explanation: Consider equal charges +q and -q separated by a distance. At the midpoint, the field due to +q points away from the positive charge, which is toward -q. The field due to -q points toward the negative charge, also in the same direction. Their magnitudes are equal and therefore add rather than cancel. The resultant field is from positive to negative, so option A is correct; option C would apply to suitable like-charge symmetry, not this dipole.

Correct answer: A. A charge of larger magnitude is represented by more lines

Explanation: Electric field lines are a graphical convention used to show the relative strength of an electric field. When the same drawing scale and line convention are used, a charge with greater magnitude is represented by a greater number of lines emerging from it if positive or terminating on it if negative. The number is not an exact physical count, but it indicates relative magnitude. Thus option A is correct; the other statements contradict this convention.

Correct answer: C. Force will be larger

Explanation: The force on a charge in an electric field is given by the vector relation F = qE. For the same charge, the magnitude is |F| = |q||E|, so increasing the field magnitude increases the force in direct proportion. The force direction is along the field for a positive charge and opposite to it for a negative charge; it does not become directionless. Therefore option C is correct, while A and B violate the relation F = qE.

Correct answer: A. South direction

Explanation: Electric field is defined as the force per unit positive test charge, E = F/q. Since the charge in this question is positive, q > 0, multiplication by q does not reverse the force direction. The force is toward the south, so the electric field at that point is also toward the south. North would be the opposite direction and east or west are perpendicular alternatives; therefore option A is correct.

Correct answer: B. North direction

Explanation: The force on a charge is F = qE. For a negative charge, q is less than zero, so the force direction is opposite to the electric-field direction. Since the observed force points south, the electric field must point north. The field is not east or west, and choosing south would incorrectly apply the positive-charge rule. Hence option B is correct.

Correct answer: A. Because field magnitude is the same everywhere

Explanation: The density or spacing of electric field lines is used to represent the relative magnitude of the field: closely spaced lines indicate a stronger field, while widely spaced lines indicate a weaker field. In a uniform electric field, both magnitude and direction remain constant throughout the region. Therefore the lines are drawn straight, parallel, and equally spaced. Equal spacing does not mean the field is zero or that its direction reverses, so option A is correct.

Correct answer: A. Zero

Explanation: The governing relation is F = qE, where F is electric force, q is the test charge, and E is the electric field. Substituting E = 0 gives F = q × 0 = 0, regardless of the small positive charge's value. Hence option A is correct. The force is not equal to the charge, and zero field cannot produce a maximum or directionless large force. This conclusion assumes electrostatic force due to the stated electric field.

Correct answer: A. Electric field is zero

Explanation: For a charge acted on by an electric field, the governing equation is F = qE. Since the charge q is explicitly non-zero and the force F is zero, division by q is valid: E = F/q = 0/q = 0. Thus option A follows directly. A very large field would generally produce a non-zero force, while changing direction and lacking units do not explain the stated zero force. Other forces are not part of the situation described.

Correct answer: A. A positive test charge is attracted toward the negative charge

Explanation: Electric-field direction is defined as the direction of the force on a positive test charge. By Coulomb's law, opposite charges attract, so a positive test charge placed near a negative point charge is pulled toward it. Therefore the field vector and the arrows on field lines point inward, making option A correct. Option B denies attraction, C reverses the rule for negative charge, and D ignores that arrows indicate direction.

Correct answer: A. Field lines are directed from positive to negative

Explanation: An electric dipole consists of equal and opposite charges, but its field is not zero everywhere. The direction of an electric field is defined by the force on a positive test charge: it is away from a positive source charge and toward a negative source charge. Consequently, dipole field-line arrows emerge from + and terminate on −, so option A is correct. Options B reverses the convention, while C and D incorrectly deny the dipole field.

Correct answer: A. 10 newton

Explanation: Use the defining relation between force and electric field: F = qE. Here q = 0.5 C and E = 20 N/C, so F = (0.5 C)(20 N/C) = 10 N. The coulomb units cancel, leaving newtons. Because the charge is positive, the force acts in the direction of the electric field. Therefore option A is correct; B, C, and D result from using the wrong arithmetic or formula.

Correct answer: C. 3 newton per coulomb

Explanation: The governing definition is E = F/q, obtained from F = qE. Substituting the given values gives E = 12 N ÷ 4 C = 3 N/C. Since the charge is positive, the electric-field direction is the same as the force direction, although the question asks only for magnitude. Therefore option C is correct. Option A multiplies instead of dividing, while B and D use incorrect numerical operations.

Correct answer: A. Direction of electric field

Explanation: An arrow on an electric field line is not merely a decorative mark; it defines the local direction of the electric field. By convention, it points in the direction of force on a positive test charge. Reversing it therefore gives an incorrect field direction and would also imply an incorrect force direction for a positive charge. Option A is correct. The reversal does not alter the charge’s mass, the paper size, or the diagram’s colour, so B, C, and D are irrelevant.

Correct answer: A. Two newtons per coulomb

Explanation: Electric field at a point is defined as force per unit positive test charge: E = F/q. Substituting the given values gives E = 4 N ÷ 2 C = 2 N/C. Therefore option A, two newtons per coulomb, is correct. Option B incorrectly uses the force without dividing by charge, option C multiplies instead of dividing, and option D would result from an incorrect numerical operation. The positive sign of the test charge does not change the magnitude calculation.

Correct answer: A. Toward the left

Explanation: The governing relation is F = qE, where electric field direction is defined as the force direction for a positive charge. Here q is negative, so the force reverses the direction of E. Since the field points to the right, the negative test charge experiences force to the left. Therefore, option A is correct; option B would apply to a positive charge, while the vertical options are unrelated.

Correct answer: A. One fourth

Explanation: For a point charge, the electric field magnitude is E = kQ/r². If the distance changes from r to 2r while Q remains unchanged, the new field is E' = kQ/(2r)² = kQ/(4r²) = E/4. Thus one-fourth of the original field remains. Option B ignores the square dependence, whereas C and D give an increase instead of the required decrease.

Correct answer: A. It becomes double

Explanation: The electric field produced by a point charge is E = kQ/r². With distance r fixed, E is directly proportional to the source-charge magnitude Q. Replacing Q by 2Q gives E' = k(2Q)/r² = 2E, so the field becomes double. Option B reverses the proportionality, option C incorrectly squares the charge change, and option D has no physical basis.

Correct answer: A. The electric field is stronger there

Explanation: Electric field lines are an illustrative representation, and their density indicates the relative strength of the field. Where the lines are closer together, more lines pass through a given area, signifying a larger electric-field magnitude. Therefore option A is correct. Close spacing does not mean the field is zero, does not exclude charge, and does not prevent the field direction from changing; those statements confuse line density with other properties.

Correct answer: A. From the direction of tangent at that point

Explanation: By definition, the electric field direction at a point is the direction in which a positive test charge would be pushed. A field line is drawn so that its tangent at any point gives this local direction. For a curved line, the tangent changes from point to point, so the normal, thickness, or colour cannot determine the field direction. Therefore option A is correct.

Correct answer: A. Force acts in the direction of the field

Explanation: The electric force is given by F = qE. For a positive charge, q is positive, so the force vector has the same direction as the electric-field vector. In a uniform field, the magnitude and direction of E are constant, although the force still depends on q. Option B applies to a negative charge; C is false unless the field or charge is zero, and D describes neither electrostatic force nor this situation. Hence A is correct.

Correct answer: A. Zero

Explanation: Electric field is a vector, so both magnitude and direction must be considered. Let the two fields be +E and −E along the same line. Their vector sum is E_net = E + (−E) = 0. Thus the fields exactly cancel. The result is not 2E or E/2, and the direction is not uncertain because a zero vector has no direction after complete cancellation.

Correct answer: A. Eight newtons per coulomb

Explanation: For collinear electric fields acting in the same direction, the vector magnitudes add directly. Taking the common direction as positive, E_net = 3 N/C + 5 N/C = 8 N/C. Therefore option A is correct. Subtraction would apply only for opposite directions, multiplication is not the rule for superposition, and the result cannot be zero when both nonzero fields point identically.

Correct answer: A. Five newtons per coulomb to the right

Explanation: Choose rightward as the positive direction. The rightward field is +8 N/C and the leftward field is −3 N/C, so E_net = +8 − 3 = +5 N/C. The positive sign means the result points to the right, the direction of the larger field. Adding the magnitudes would ignore direction, while zero would require equal opposite fields; hence option A is correct.