An electron is emitted from a nucleus- State the effect this has on the charge of the nucleus
2 answers:
An electron is emitted from a nucleus- State the effect this has on the charge of the nucleus
Answer:
an electron orbiting around the nucleus combines with a nuclear proton to produce a neutron, which remains in the nucleus, and a neutrino, which is emitted. As in positron emission, the nuclear positive charge and hence the atomic number decreases by one unit, and the mass number remains the same.
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<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>
When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:
To maintain constant, if Q increases V also increases.
On the other hand, the potential energy can be expressed as:
In conclusion, as Q increases the potential energy also increases.
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.