Free Electric Field Intensity MCQs with Answers

1. Electric field intensity at a point is defined as

• 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 charge would be pushed. Work done per unit charge is potential, a scalar, and confusing the two is the most common error in this chapter. The test charge must be small enough not to disturb the field it is measuring.

2. The electric field intensity at a distance r from a point charge q is

• 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 points radially outwards for a positive charge. Field lines drawn from such a charge spread out into space, and their decreasing density with distance is a picture of the same falling intensity. For a uniform field, as between parallel plates, E is constant instead.

3. Two identical charges are placed a fixed distance apart. The point where the resultant field is zero lies

• 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. If the charges were unlike, no point between them could give cancellation because both fields would point the same way, and the null point would lie outside the pair on the side of the smaller charge. Working out which case applies is the first step in any such question.

4. The electric field at a point due to two equal and opposite charges is 100 N/C. If the magnitude of each charge is doubled then the electric field at that point becomes:

• 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 doubling both charges doubles each contribution and doubles the resultant, which by that reasoning gives 200 N/C. The key's figure of 400 corresponds to treating the effect as multiplying by four, so this is a question where the official answer and the standard proportional argument diverge.

5. 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 that density and rises by the same 25 per cent. Notably the field does not depend on distance from the sheet at all, which is what makes the infinite sheet such a useful idealisation. A point charge behaves quite differently, falling off as the inverse square.

6. For two equal positive charges, the electric field is weakest:

• 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 is the weakest the field can be anywhere at finite distance. Close to either charge the field is strongest. The field also tends to zero at infinity, but the question is asking for the null point produced by the pair.