This table shows experiments done by five scientists. Scientist Experiment Kyle cut a bar of gold in half Katie burned paper int
2 answers:
You might be interested in
Answer:
a) the capacitance is the same for both capacitors.
b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q.
c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q
d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q
Explanation:
a) The capacitance of a capacitor, by definition, is as follows:
Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be expressed as follows:
C = ε*A / d
As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.
b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.
If this proportion must remain the same, and one of the capacitors has the double of the charge than the other, the potential difference must be the double also.
c) Applying Gauss' law, to the surface of one of the plates, and assuming a constant surface charge density σ, it can be showed that the electric field can be calculated as follows:
E*A = Q/ε₀ as σ=Q/A
⇒ E = σ/ε₀
As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.
c) The electric potential energy, stored between plates of a capacitor, can be written as follows:
If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.
Answer:
Q = +1.4 pC
Explanation:
- The capacitance of any capacitor, by definition, is as follows:
-
- Applying Gauss'Law and the definition of electric potential, it can be showed that the capacitance of a parallel-plate capacitor can be expressed as follows:
-
- Where εr is the dielectric constant of the material that fills the space between the plates, A is the area of one of the plates, and d, is the separation between them.
- Replacing by the givens in (2) we can find the value of the capacitance C, as follows:
- Replacing the values of C and V in (1), we can solve for Q, as follows:
- As the outer surface is at a higher potential that the inside surface, the charge on it must be positive, and is equal to +1.4 pC.