Question 2
2
4 marks
Question 2(a)
2(a)
State Coulomb's law.
Easystructured2 marks
Answer
(electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1
Question 2(b)
2(b)
Positronium is a system in which an electron and a positron orbit, with the same period, around their common centre of mass, as shown in Fig. 2.1. The radius r of the orbit of both particles is \(1.59 \times 10^{-10} \mathrm{~m}\).
structured2 marks
Question 2(b)(ii)
2(b)(ii)
Show that the magnitude of the electric force between the electron and the positron is \(2.28 \times 10^{-9} \mathrm{~N}\).
Mediumstructured2 marks
Answer
\(F=e^{2} / 4 \pi \varepsilon_{0} X^{2}\) C1 \[ \begin{aligned} =\left(1.60 \times 10^{-19}\right)^{2} /\left[4 \pi \times 8.85 \times 10^{-12} \times\left(2 \times 1.59 \times 10^{-10}\right)^{2}\right] =2.28 \times 10^{-9} \mathrm{~N} \end{aligned} \] A1
Question 4
4
A helium nucleus contains two protons. In a model of the helium nucleus, each proton is considered to be a charged point mass. The separation of these point masses is assumed to be \(2.0 \times 10^{-15} \mathrm{~m}\).
structured2 marks
Question 4(a)
4(a)
For the two protons in this model, calculate
structured2 marks
Question 4(a)(i)
4(a)(i)
the electrostatic force, electrostatic force = N
Mediumstructured2 marks
Answer
aligned _E & =Q_1 Q_2 / 4 _0 r^2 & =8.99 x 10^9 x (1.6 x 10^-19)^2 /(2.0 x 10^-15)^2 & =58 N aligned
Question 4
4
2 marks
Question 4(d)
4(d)
By reference to Fig. 4.2, state and explain
structured2 marks
Question 4(d)(i)
4(d)(i)
whether the two charges have the same, or opposite, sign,
Easystructured2 marks
Answer
work is got out as x decreases M1 so opposite sign A1 [2]
Question 4
4
2 marks
Question 4(a)
4(a)
State Coulomb's law.
Easystructured2 marks
Answer
force proportional to product of charges and inversely proportional to the square of the separation B1 force between point charges B1
Question 4
4
Two point charges A and B each have a charge of \(+6.4 \times 10^{-19} \mathrm{C}\). They are separated in a vacuum by a distance of \(12.0 \mu \mathrm{~m}\), as shown in Fig. 4.1. Points P and Q are situated on the line A B. Point P is \(3.0 \mu \mathrm{~m}\) from charge A and point Q is \(3.0 \mu \mathrm{~m}\) from charge B.
structured2 marks
Question 4(a)
4(a)
Calculate the force of repulsion between the charges A and B . force =
Mediumstructured2 marks
Answer
force \(=q_{1} q_{2} / 4 \pi \varepsilon_{0} x^{2} \quad\) C1 \(=\left(6.4 \times 10^{-19}\right)^{2} /\left(4 \pi \times 8.85 \times 10^{-12} \times\left\{12 \times 10^{-6}\right\}^{2}\right) \quad \mathrm{C} 1\) \(=2.56 \times 10^{-17} \mathrm{~N}\) A1
Question 5
5
3 marks
Question 5(b)
5(b)
An isolated solid metal sphere is positively charged. The variation of the potential V with distance x from the centre of the sphere is shown in Fig. 5.1. Use Fig. 5.1 to suggest
structured3 marks
Question 5(b)(ii)
5(b)(ii)
that the charge on the sphere behaves as if it were a point charge.
Mediumstructured3 marks
Answer
for point charge, V x is constant B1 co-ordinates clear and determines two values of V x at least 4 cm apart M1 conclusion made clear A1 [3]
Question 4
4
An \(\alpha\)-particle and a proton are at rest a distance \(20 \mu \mathrm{~m}\) apart in a vacuum, as illustrated in Fig. 4.1.
structured4 marks
Question 4(a)
4(a)
4 marks
Question 4(a)(i)
4(a)(i)
State Coulomb's law.
Easystructured2 marks
Answer
force proportional to product of (two) charges and inversely proportional to square of separation reference to point charges
Question 4(a)(ii)
4(a)(ii)
The \(\alpha\)-particle and the proton may be considered to be point charges. Calculate the electric force between the \(\alpha\)-particle and the proton. force = N
Mediumstructured2 marks
Answer
\(F=2 \times\left(1.6 \times 10^{-19}\right)^{2} /\left\{4 \pi \times 8.85 \times 10^{-12} \times\left(20 \times 10^{-6}\right)^{2}\right\}\)
Question 4
4
Two small charged metal spheres A and B are situated in a vacuum. The distance between the centres of the spheres is 12.0 cm , as shown in Fig. 4.1. The charge on each sphere may be assumed to be a point charge at the centre of the sphere. Point P is a movable point that lies on the line joining the centres of the spheres and is distance x from the centre of sphere A . The variation with distance x of the electric field strength E at point P is shown in Fig. 4.2.
structured2 marks
Question 4(a)
4(a)
State the evidence provided by Fig. 4.2 for the statements that
structured2 marks
Question 4(a)(ii)
4(a)(ii)
the charges on the spheres are either both positive or both negative.
Mediumstructured2 marks
Answer
either field strength is zero or M1 A1 [2]
Question 4
4
An \(\alpha\)-particle and a proton are at rest a distance \(20 \mu \mathrm{~m}\) apart in a vacuum, as illustrated in Fig. 4.1.
structured4 marks
Question 4(a)
4(a)
4 marks
Question 4(a)(i)
4(a)(i)
State Coulomb's law.
Easystructured2 marks
Answer
force proportional to product of (two) charges and inversely proportional to square of separation reference to point charges A1
Question 4(a)(ii)
4(a)(ii)
The \(\alpha\)-particle and the proton may be considered to be point charges. Calculate the electric force between the \(\alpha\)-particle and the proton. force = N
Mediumstructured2 marks
Answer
\(F=2 \times\left(1.6 \times 10^{-19}\right)^{2} /\left\{4 \pi \times 8.85 \times 10^{-12} \times\left(20 \times 10^{-6}\right)^{2}\right\} \quad \mathrm{C} 1 =1.15 \times 10^{-18} \mathrm{~N}\)
Question 4
4
2 marks
Question 4(a)
4(a)
State Coulomb's law.
Easystructured2 marks
Answer
(electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1