What would happen when two or more capacitors are connected in series across a potential difference, then
2 answers:
Answer:
The Current Iₜ = I₁ = I₂ = I₃
Charge Qₜ = Q₁ = Q₂ = Q₃
Potential difference Vₜ = V₁ + V₂ + V₃
Capacitance 1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃
Explanation:
According to the attached image;
For series arrangements of capacitors, the current flowing through each of the capacitors are the same and equal to the total current flowing through the circuit;
Iₜ = I₁ = I₂ = I₃
Also the charge storage by each capacitor is equal;
Qₜ = Q₁ = Q₂ = Q₃
The potential difference across each of the capacitors sum up to the total voltage across the circuit;
Vₜ = V₁ + V₂ + V₃
The reciprocal of the total capacitance equals the sum of the reciprocal of capacitances of each of the capacitors;
1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃
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Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)
Answer:
d= 0.46 m
Explanation:
The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.
For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:
V =
We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.
If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.
So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:
V = +
= 0
⇒:
Replacing by the values of q1, q2, and k, and solving for x, we get:
⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.