Electric Fields tough question 1 (From Note)

In a simplified model, a uranium nucleus is a sphere of radius 8.0 × 10⁻¹⁵ m. The nucleus contains 92 protons (and rather more neutrons). The charge on a proton is 1.6 × 10⁻¹⁹ C. It can be assumed that the charge of these protons acts as if it were all concentrated at the centre of the nucleus. The nucleus releases an α particle containing two protons (and two neutrons) at the surface of the nucleus. Calculate

a)  the electric field strength at the surface of the nucleus before emission of the α-particle,

b)  the electric force on the α-particle at the surface of the nucleus,

c)  the electric potential at the surface of the nucleus before emission of the α-particle,

d)  the electric potential energy of the α-particle when it is at the surface of the nucleus.

Solution:

Capacitor Question 3

(a) A 470 μF capacitor is connected to a 20 V supply. Calculate the charge stored on one plate of the capacitor.
(b) The capacitor in a is now disconnected from the supply and connected to an uncharged 470 μF capacitor.
(i) Explain why the capacitors are in parallel, rather than series, and why the total charge stored by the combination must be the same as the answer to (a).
(ii) Calculate the capacitance of the combination.
(iii) Calculate the potential difference across each capacitor.
(iv) Calculate the charge stored on one plate of each capacitor.

Solution:

a)

Q = CV

= (470 x 10^-6) x (20)

= 9.4 mC

bi)

When capacitors are connected in series, the total capacitance is less than any one of the series capacitors’ individual capacitances. If two or more capacitors are connected in series, the overall effect is that of a single (equivalent) capacitor having the sum total of the plate spacings of the individual capacitors. As we’ve just seen, an increase in plate spacing, with all other factors unchanged, results in decreased capacitance.

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors’ capacitances. If two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor having the sum total of the plate areas of the individual capacitors. As we’ve just seen, an increase in plate area, with all other factors unchanged, results in increased capacitance.

The total charge stored by the combination must be the same as the answer to (a) is because the total charge of a capacitor is equal to the total charge of both capacitors.

bii) 

C = C1 + C2

= 470 uF + 470 uF

= 940 uF

biii)

Q ∝ V
1Q = 20V

0.5Q = 10V

biv)

Q1 = (C1) (V) 

= (470 uF)(10)

= 4.7 mC

Q2 = (C1) (V) 

= (470 uF)(10)

= 4.7 mC

Reference:

https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitors-parallel

https://www.allaboutcircuits.com/textbook/direct-current/chpt-13/series-and-parallel-capacitors/

Electric Fields Tough Question 1

Two +30 μC charges are placed on a straight line 0.40 m apart. A +0.5 μC charge is to be moved a distance of 0.10 m along the line from a point midway between the charges. How much work must be done?

Solution:

Examination Style Questions 7 from Oscillations note

The apparatus of Fig. 13.16 is used to demonstrate forced vibrations and resonance. A 50 g mass is suspended from the spring, which has a spring constant of 7.9 N m⁻¹.

a)  Calculate the resonant frequency f₀ of the system.

b)  A student suggests that resonance should also be observed at a frequency of 2 f₀. Discuss this suggestion.

Solution:

a)

b) If the driving frequency is increased further, the amplitude of oscillation of the mass decreases.

Reference: Examination Style Questions 7 from note

Examination Style Questions 5 from Oscillations note

One particle oscillating in simple harmonic motion has ten times the total energy of another particle, but the frequencies and masses are the same. Calculate the ratio of the amplitudes of the two motions.

Solution:

Reference: Examination Style Questions 5 from Oscillations note

Examination Style Questions 2 from Oscillations note

A particle is oscillating in simple harmonic motion with frequency 50 Hz and amplitude 15 mm. Calculate the speed when the displacement from the equilibrium position is 12mm.

Solution:

Reference: Examination Style Questions 2 from note

Examination Style Questions 1 from Oscillations note

A particle is oscillating in simple harmonic motion with period 4.5ms and amplitude 3.0cm. At time t = 0, the particle is at the equilibrium position. Calculate, for this particle: 

a)the frequency,
b)the angular frequency,
c)the maximum speed,
d)the magnitude of the maximum acceleration,
e)the speed at time t = 1.0 ms.

Solution:

Reference: Examination Style Questions 1 from Oscillations note