# Potential difference equation

3 questions about Potential difference equation

1. An electric charge is moved in a homogeneous electric field with a force of 2√3 N a distance of 20 cm. If the direction of the force is at an angle of 30o to the displacement of the electric charge, what is the difference in the electric potential energy at the initial and final positions of the electric charge.

Known:

Force (F) = 2√3 N

Distance (s) = 20 cm = 0.2 m

Angle (θ) = 30o

Wanted: Electric potential difference

Solution:

The work of transferring charge +q from a to b is equal to the electric potential energy difference at points a and b.

ΔEP = ΔW

If the direction of the force F with respect to the direction of charge transfer +q is angled θ, then the work of charge transfer +q from a to b is:

ΔW = F Δs cos θ = (23)(0,2)(cos 30) = (0,43)(0,53) = (0,2)(3) = 0,6 Joule

2. If the charge and capacity of the capacitor are known to be 5 µC and 20 µF, respectively, determine the potential difference of the capacitors.

Known:

Electric charge (q) = 5 µC = 5 x 10-6 C

Capacitor capacity (C) = 20 µF = 20 x 10-6 F

Wanted: Capacitor potential difference (V)

Solution:

The formula for capacitor potential difference, capacitor capacity and charge:

V = q / C = (5 x 10-6) / 20 x 10-6 = 5/20 = 0,25 Volt

3. Two point charges QA = -4 µC and QB = 8 µC are 16 cm apart. Determine the electric potential at a point halfway between the two charges.

Known:

Electrical charge 1 (QA) = 4 x 10-6 C

Electrical charge 2 (QB) = 8 x 10-6 C

Distance (r) = 8 cm = 0,08 m

Coulomb constant (k) = 9 × 109 Nm2/C2

Wanted: The electric potential at C (VC)

Solution:

VA = k Q / r = (9 × 109)(4 x 10-6) / 0,08 = (36 × 103) / 0,08 = -450 Volt

VB = k Q / r = (9 × 109)(8 x 10-6) / 0,08 = (72 × 103) / 0,08 = 900 Volt

The electric potential at C (VC) = 900 – 450 = 450 Volt