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 30^o 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) = 30^o

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 = (2√3)(0,2)(cos 30) = (0,4√3)(0,5√3) = (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 (Q_A) = 4 x 10^-6 C

Electrical charge 2 (Q_B) = 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 (V_C)

Solution:

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

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

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