Electric Field | MCQs

Explanation

The energy stored in a capacitor is given by U = (1/2)CV². If the potential difference (V) is doubled:

U ∝ V²

U' ∝ (2V)² = 4V²

So, the energy stored will increase by a factor of 4.

Explanation

Given the formula and calculation:

F = qE

= (1.6 × 10⁻¹⁹ C) × (1.8 × 10⁻⁸ N/C)

= 2.88 × 10⁻¹¹ N

Explanation

Electric field intensity is defined as force per unit charge (N/C) or potential difference per unit distance (V/m).

So, V/m is an equivalent.

It is commonly used unit for electric field intensity.

Explanation

Energy density is defined as energy per unit volume, so the SI unit is joules per cubic meter (J/m³).

It applies to electric fields where

𝑢 = 1/2𝜀𝐸², with 𝑢 being energy density.

Explanation

Electric fields can be detected by electrons, which are charged particles that respond to the force exerted by the electric field.

Electrons will accelerate or decelerate in the presence of an electric field, allowing for the detection of the field.

Explanation

The work done on a unit charge against an electric field intensity is called electric potential.

It is also known as voltage or potential difference.

It is measured in volts (V) and represents the energy transferred per unit charge.

Electric potential (V) = Work done (W) / Charge (Q)

V = W / Q

Explanation

The direction of the electric force (F) and electric field intensity (E) are parallel to each other.

The electric field intensity is a vector quantity that points in the direction of the force that would be experienced by a test charge at a given point.