Free Electric Field Intensity MCQs with Answers
30 Electric Field Intensity MCQs from Physics, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Electric field intensity is the force acting per unit positive test charge, with direction set by the force on that charge. Work includes fields of point charges, vector superposition, field lines and the relation between field and electric potential. Field intensity must be distinguished from electric force and potential.
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- A. the force per unit positive charge placed at that point
- B. the work done in bringing a charge to that point
- C. the charge per unit area
- D. the potential at that point
Explanation: Field intensity is a vector measured in newtons per coulomb, equivalently volts per metre, and it points in the direction a positive test…
Correct answer: the force per unit positive charge placed at that point- A. proportional to q and to r
- B. proportional to q and inversely proportional to r squared
- C. independent of r
- D. proportional to r squared
Explanation: Dividing the Coulomb force by the test charge leaves E equal to kq over r squared, so the field also obeys an inverse square law and…
Correct answer: proportional to q and inversely proportional to r squared- A. midway between them
- B. at one of the charges
- C. outside the pair, beyond the smaller charge
- D. nowhere, since the field is never zero
Explanation: For like charges of equal magnitude the two fields are equal and opposite at the midpoint, so they cancel there.
Correct answer: midway between them- A. 50 N/C
- B. 100 N/C
- C. 200 N/C
- D. 400 N/C
Explanation: This is the option marked correct in the official UHS answer key. Field intensity is directly proportional to the charge producing it, so…
Correct answer: 400 N/C5. If the surface charge density of an infinite sheet increases by 25%, the electric field intensity:
- A. Increases by 25%
- B. Increases by 50%
- C. Decreases by 25%
- D. Remains the same
Explanation: The field of an infinite charged sheet is the surface charge density divided by twice the permittivity, so it is directly proportional to…
Correct answer: Increases by 25%- A. Midway between them
- B. Along the perpendicular bisector
- C. Close to either charge
- D. At infinity
Explanation: At the midpoint the two fields are equal in magnitude and opposite in direction, so they cancel exactly and the resultant is zero, which…
Correct answer: Midway between them- A. 9 x 10^5 N / C
- B. 3 x 10^3 N / C
- C. 4.5 x 10^5 N / C
- D. 9 x 10^-5 N / C
Explanation: The electric field E due to a point charge q at a distance r from the charge is given by Coulomb's law: E=kq /r2where:k is Coulomb's…
Correct answer: 9 x 10^5 N / C- A. 3 x 10^5N/C
- B. 4.5 x 10^5N/C
- C. 9 x 10^5N/C
- D. 3 x 10^3N/C
Explanation: The electric field E due to a point charge q at a distance r from the charge is given by Coulomb's law: E=kq /r2where:k is Coulomb's…
Correct answer: 9 x 10^5N/C- A. 4E
- B. E/2
- C. E/4
- D. 2E
Explanation: Between two parallel plates, the electric field strength is given by:E = V/dIf the distance is halved, electric field strength is doubled…
Correct answer: 2E- A. 1.82 x 10^-15 N
- B. 3.00 x 10^-16 N
- C. 1.92 x 10^-15 N
- D. 3.00 x 10^-15 N
- E. 5.00 x 10^-15 N
Explanation: To solve this problem, we first calculate the electric field (E) between the parallel plates using the formula E = V/d, where V is the…
Correct answer: 1.92 x 10^-15 N- A. Newton meter
- B. Newton / meter
- C. Newton coulomb
- D. Newton / coulomb
Explanation: The formula for the electric field strength can be written as:E=Force/ChargeSubstituting the corresponding SI units for these two…
Correct answer: Newton / coulomb- A. Electric field intensity
- B. Electric flux
- C. Magnetic intensity
- D. Magnetic flux
Explanation: W = FdF = qEW = qEdW/q = V = EdV/d = EThe potential gradient is represented by -Δd/Δr.
Correct answer: Electric field intensity- A. 4 cm/s2
- B. 2 cm/s^2
- C. 40 cm/s2
- D. 20 cm/s2
Explanation: 2x10-6 x 20 / 2x10-3 = 0.02 m/s2 x100 = 2cm/s2
Correct answer: 2 cm/s^2- A. E
- B. 2E
- C. 4E
- D. 8E
Explanation: Electric field is calculated by the formula = Voltage / Distance so doubling the separation of plates decreases the electric field by 2…
Correct answer: 2E- A. no direction.
- B. direction along the area of surface.
- C. direction along the electrostatic force.
- D. direction making an angle of 45o with force.
Explanation: Electric field intensity is defined as the electrostatic force experienced by a unit positive charge.
Correct answer: direction along the electrostatic force.- A. Zero between the plates
- B. Stronger near the plate P
- C. Stronger near the plate Q
- D. Uniform between the plates
Explanation: The correct answer is that the intensity of the electric field is uniform between the plates.
Correct answer: Uniform between the plates- A. 8.0 x 10^-22 N
- B. 3.2 x 10^-17 N
- C. 2.0 x 10^2 N
- D. 1.2 x 10^21 N
Explanation: The correct answer is Option B: 3.2 × 10-17 N. To calculate the electrostatic force on an electron, we use the formula F = qE, where q is…
Correct answer: 3.2 x 10^-17 N- A. qE/m
- B. qEm
- C. q/Em
- D. m/qE
Explanation: The equation for the acceleration of a charged particle in a uniform electric field isa = q E / m ,wherea is the acceleration,q is the…
Correct answer: qE/m- A. Force per unit mass
- B. Force per unit tesla
- C. Force per unit charge
- D. Force per unit watt
Explanation: A measure of the force exerted by one charged body on another. Imaginary lines of force or electric field lines originate (by convention)…
Correct answer: Force per unit charge- A. E = m*g/ q
- B. E = m*g/ q*v
- C. E = m*g/ q*v*B
- D. E = q / m*g
Explanation: The formula for Electric Field is given by;E = F/qWhere, according to Newton's second law of motion, F = maE = ma/qIn the problem, the…
Correct answer: E = m*g/ qElectric Field Intensity MCQs: common questions
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There are 30 Electric Field Intensity MCQs in the Physics bank, shown 20 to a page with the correct answer and an explanation on each.
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