Changes in speed are solved related to kinetic energy or momentum. We don't have collisions, we can't apply momentum.

We are going to apply the conservation of energy because kinetic energy and electric potential energy are related using the conservation of energy.

A charge q in a position where an electric potential is V has electric potential energy qV

Positive charges have positive electric potential energy; negative charges have negative electric potential energy. We need to include the sign of charge q in U = qV.

Interpretation of U = qV

U becomes more positive as V increases (toward high potential) for positive charges. ΔU = U_f - U_i is positive; potential energy increases; kinetic energy decreases

U becomes more negative as V increases (toward high potential) for negative charges. ΔU = U_f - U_i is negative; potential energy decreases; kinetic energy increases.

Using conservation of energy:

K₀ + U₀ = K_f + U_f

(1/2)mv₀² + qV₀ = (1/2)mv_f²+ qV_f

Our target is v₀:

v₀² = 2/m[(1/2)mv_f² + qV_f - qV₀] = v_f² + (2q/m)(V_f- V₀) = v_f² + (2q/m)ΔV

ΔV is unknown.

The relationship between electric potential difference ΔV, electric field E, and plate separation d for a capacitor is:

E = ΔV/d

ΔV = Ed = (4.0 × 10⁷ V/m)(1.0 mm × 1m/1000 mm) = 40000V

We need to determine the sign of ΔV = V_f - V_i

V_f is at the positive plate (high potential); V_i is at the negative plate (low potential)

ΔV = V_f - V_i= (high potential) - (low potential) = +ve

(2q/m)ΔV: q and ΔV are positive; kinetic energy must increase.

v₀² = 2/m[(1/2)mv_f² + qV_f- qV₀] = v_f² + (2q/m)(V_f - V₀) = v_f² + (2q/m)ΔV

v₀= √[v_f² + (2q/m)ΔV] = √[(3.2 × 10⁶ m/s)² + (2 × 1.6 × 10^-19 C / (1.67 × 10^-27 kg))(40,000V)] = 4.2 × 10⁶ m/s

Relationships Between Force, Field, Energy, Potential Practice Problems