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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Thursday, November 30, 2017

An isolated metal sphere is charged to a potential V. The charge on the sphere is q. The charge on the sphere may be considered to act as a point charge at the centre of the sphere.




Question 2
(a) Define electric potential at a point.                                                [2]


(b) An isolated metal sphere is charged to a potential V. The charge on the sphere is q.
The charge on the sphere may be considered to act as a point charge at the centre of the sphere.

The variation with potential V of the charge q on the sphere is shown in Fig. 5.1.


Fig. 5.1

Use Fig. 5.1 to determine
(i) the radius of the sphere,                                                                 [2]
(ii) the energy required to increase the potential of the sphere from zero to 24 kV. [3]


(c) The sphere in (b) discharges by causing sparks when the electric field strength at the surface of the sphere is greater than 2.0 × 106 V m-1.
Use your answer in (b)(i) to calculate the maximum potential to which the sphere can be
charged. [3]





Reference: Past Exam Paper – November 2014 Paper 43 Q5





Solution:
(a) The electric potential at a point is defined as the work done / energy in moving unit positive charge from infinity (to the point)


(b) (i)
V = q / 4πε0r

{from the graph,} at 16 kV, q = 3.0 × 10–8 C

{r = q / 4πε0V}
r = (3.0 × 10–8) / (4π × 8.85 × 10–12 × 16 × 103)                    C1
r = 1.69 × 10–2 m        (allow 2 s.f.)                                        A1


(ii)
The energy is represented by the area ‘below’ line
{Energy = area of triangle between V = 0 kV to V = 24 kV}

{when V = 24 kV, q = 4.5 × 10–8 C}

Energy = ½ qV
Energy = ½ × 24 × 103 × 4.5 × 10–8
Energy = 5.4 × 10–4 J                        


(c)
since V = q / 4πε0r      and      E = q / 4πε0r2              giving Er = V   (or V = E/r)

{The maximum electric fields strength at which the sphere can be charged is E = 2.0 × 106 V m-1.}
2.0 × 106 × 1.7 × 10–2 = V
V = 3.4 × 104 V

Wednesday, November 29, 2017

A student measures the time T for one complete oscillation of a pendulum of length l. Her results are shown in the table.




Question 5
A student measures the time T for one complete oscillation of a pendulum of length l.

Her results are shown in the table.

l / m                             T / s
0.420 ± 0.001                         1.3 ± 0.1

She uses the formula
T = 2π (l/g)

to calculate the acceleration of free fall g.

What is the best estimate of the percentage uncertainty in the value of g?
A 0.02%                      B 4%               C 8%               D 16%





Reference: Past Exam Paper – November 2015 Paper 12 Q5





Solution:
Answer: D. 

T = 2π √(l/g)

T2 = 4π2 l / g                giving  g = 4π2 l / T2


Percentage uncertainty in g = (Δg / g) × 100%

Δg / g = (Δl / l) + 2(ΔT / T)
Δg / g = (0.001 / 0.420) + 2(0.1 / 1.3)
Δg / g = 0.156

Percentage uncertainty in g = 0.156 × 100 % = 15.6 % ≈ 16%

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