What would happen when two or more capacitors are connected in parallel across a potential difference?

The Current Iₜ = I₁ + I₂ + I₃

Charge Qₜ = Q₁ + Q₂ + Q₃

Potential difference Vₜ = V₁ = V₂ = V₃

The total capacitance Cₜ = C₁ + C₂ + C₃

Explanation:

According to the attached image;

For parallel arrangements of capacitors, the current flowing through each of the capacitors sums up to the total current flowing through the circuit;

Iₜ = I₁ + I₂ + I₃

Also the charge storage by each capacitor sums up to give the total charge stored;

Qₜ = Q₁ + Q₂ + Q₃

The potential difference across each of the capacitors are the same and equal to the total voltage across the circuit;

Vₜ = V₁ = V₂ = V₃

The total capacitance equals the sum of the capacitances of each of the capacitors;

Cₜ = C₁ + C₂ + C₃

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j