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Question

An electric field of 100 N/C points to the right. What is the electric force on a +2 C charge placed in this field?

This question was previously asked in
RRB Technician 2026 Grade 3 Question Paper (7-Oct-2026) (Shift 3)
The correct answer is

200 N to the right

The force on a charge in an electric field is \(\vec{F} = q\vec{E}\).

Magnitude: \(F = qE = 2 \times 100 = 200\ \text{N}\).

For a positive charge, the force acts in the same direction as the field. The field points to the right, so the force also points to the right.

Hence, the answer is 200 N to the right.

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Important Questions from Electric Fields and Gauss' Law

  1. The surface charge density of a thin spherical shell placed in an air medium is 88.54 c/m2 The intensity of the electric field measured 12 mm outside the shell from the centre of the shell is 5.625 × 101 2 N/C. The thin spherical shell has a radius of: 

  2. The electric flux passing through a surface of area A = 8j m2 in an electric field vector E = 2i + 3j - 4k V/m (bold is for vectors) is:

  3. Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho(r) = Cr^2$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance, '$a$' from the centre. If $A$ and $B$ do not experience any force, then:
  4. Two point charges $q_1 \left( {\sqrt {10} {\rm{\mu C}}} \right)$ and $q_2(-18\sqrt{2} {\rm{\mu C}})$ are placed on the x-axis at $x = 0$ m and $x = 4$ m respectively. The electric field (in V/m) at a point $(1, 3)$ m is,
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  5. The electric field between the two plates of a parallel plate (area = 5.0 × 10-2m2) capacitor is given by E = (2.0 × 105 – 4.0 × 104 t) V/m where t is in second. What is the magnitude of the displacement current between the plates? (ε0 = 8.85 × 10-12 F/m)

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