Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference
Put the value into the formula
The potential on the second plate
(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance
Put the value into the formula
Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Answer:
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Explanation:
Answer:
The most common oxidation numbers for a given element
Answer:
Explanation:
The electric field is defined as the electric force per unit of charge, this is:
.
The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as
.
By substitution we get that
Now, letting 
be the electric field at point A, letting 
be the electric field at point B, and letting R be the distance from the charge to A:
.
The ration of the electric fields is
This means that at half the distance, the electric field is four times stronger.
Left. Opposite of the direction the box is pushed.
