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A-Level CAIE Physics 18 4 Electric Field Of A Point Charge Question Bank

Practice A-Level CAIE Physics 18 4 Electric Field Of A Point Charge questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.

9 matching questions ยท Open interactive library

Question 1

1

2 marks

Question 1(c)

1(c)

Assume that the Earth is a uniform conducting sphere of mass \(5.98 \times 10^{24} \mathrm{~kg}\). The surface of the Earth carries a charge of \(-4.80 \times 10^{5} \mathrm{C}\) that is evenly distributed.

structured2 marks

Question 1(c)(i)

1(c)(i)

Use the information in (b) to determine the electric field strength at the surface of the Earth. Give a unit with your answer. electric field strength = unit

Hardstructured2 marks

Answer

\[ \begin{aligned} E =\alpha g Q / M =\left(1.35 \times 10^{20} \times 9.81 \times 4.80 \times 10^{5}\right) /\left(5.98 \times 10^{24}\right) \end{aligned} \] C1 \(=106 \mathrm{~N} \mathrm{C}^{-1}\) or \(106 \mathrm{~V} \mathrm{~m}^{-1}\) A1

Question 4

4

2 marks

Question 4(d)

4(d)

By reference to Fig. 4.2, state and explain

structured2 marks

Question 4(d)(ii)

4(d)(ii)

the effect, if any, on the shape of the graph of doubling the charge on particle P .

Mediumstructured2 marks

Answer

energy would be doubled B1 gradient would be increased B1

Question 5

5

An isolated solid metal sphere of radius r is given a positive charge. The distance from the centre of the sphere is x.

structured3 marks

Question 5(b)

5(b)

The electric field strength at the surface of the sphere is \(E_{0}\). On the axes of Fig.5.2, sketch a graph to show the variation with distance x of the electric field strength due to the charged sphere, for values of x from x=0 to x=4 r.

Mediumstructured3 marks

Answer

graph: straight line at E=0 from x=0 to \(x=r \quad\) B1 curve with decreasing gradient from ( \(r, E_{0}\) ) M1 passing through \(\left(2 r, 1 / 4 E_{0}\right)\) A1 (for 3rd mark line must be drawn to x=4 r and must not touch x-axis)

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

structured1 marks

Question 5(b)(i)

5(b)(i)

why the radius of the sphere cannot be greater than 1.0 cm ,

Mediumstructured1 marks

Answer

inside the sphere, the potential would be constant B1

Question 5(c)

5(c)

Assuming that the charge on the sphere does behave as a point charge, use data from Fig. 5.1 to determine the charge on the sphere. charge = C

Mediumstructured2 marks

Answer

\(q=4 \pi \varepsilon_{0} V x\) \(q=4 \pi \times 8.85 \times 10^{-12} \times 180 \times 1.0 \times 10^{-2} \quad\) M1 \(=2.0 \times 10^{-10} \mathrm{C}\) A1

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.

structured0 marks

Question 4(b)

4(b)

0 marks

Question 4(b)(ii)

4(b)(ii)

A point P is distance x from the \(\alpha\)-particle along the line joining the \(\alpha\)-particle to the proton (see Fig. 4.1). The variation with distance x of the electric field strength \(E_{\alpha}\) due to the \(\alpha\)-particle alone is shown in Fig. 4.2. The variation with distance x of the electric field strength \(E_{\mathrm{p}}\) due to the proton alone is also shown in Fig. 4.2.

Mediumstructured0 marks

Answer

1. electric field is a vector quantity electric fields are in opposite directions charges repel Any two of the above, 1 each 2. graph: line always between given lines crosses x-axis between \(11.0 \mu \mathrm{~m}\) and \(12.3 \mu \mathrm{~m}\) reasonable shape for curve

Question 4

4

A charged point mass is situated in a vacuum. A proton travels directly towards the mass, as illustrated in Fig. 4.1. When the separation of the mass and the proton is r, the electric potential energy of the system is \(U_{p}\). The variation with r of the potential energy \(U_{\mathrm{p}}\) is shown in Fig. 4.2.

structured2 marks

Question 4(a)

4(a)

2 marks

Question 4(a)(i)

4(a)(i)

Use Fig. 4.2 to state and explain whether the mass is charged positively or negatively.

Mediumstructured2 marks

Answer

as r decreases, energy decreases/work got out (due to) M1 attraction so point mass is negatively charged A1

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.

structured0 marks

Question 4(b)

4(b)

0 marks

Question 4(b)(ii)

4(b)(ii)

A point P is distance x from the \(\alpha\)-particle along the line joining the \(\alpha\)-particle to the proton (see Fig. 4.1). The variation with distance x of the electric field strength \(E_{\alpha}\) due to the \(\alpha\)-particle alone is shown in Fig. 4.2. The variation with distance x of the electric field strength \(E_{\mathrm{p}}\) due to the proton alone is also shown in Fig. 4.2.

Mediumstructured0 marks

Answer

1. electric field is a vector quantity electric fields are in opposite directions charges repel Any two of the above, 1 each B2 2. graph: line always between given lines M1 crosses x-axis between \(11.0 \mu \mathrm{~m}\) and \(12.3 \mu \mathrm{~m} \quad\) A1 reasonable shape for curve A1

Question 5

5

2 marks

Question 5(b)

5(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. Use Fig. 5.1 to determine

structured2 marks

Question 5(b)(i)

5(b)(i)

the radius of the sphere, radius = m

Mediumstructured2 marks

Answer

\(V=q / 4 \pi \varepsilon_{0} r\) at \(16 \mathrm{kV}, q=3.0 \times 10^{-8} \mathrm{C}\) (allow any answer which rounds to \(1.7 \times 10^{-2}\) )

Question 5(c)

5(c)

The sphere in (b) discharges by causing sparks when the electric field strength at the surface of the sphere is greater than \(2.0 \times 10^{6} \mathrm{Vm}^{-1}\). Use your answer in (b)(i) to calculate the maximum potential to which the sphere can be charged. potential = V

Hardstructured0 marks

Answer

\(V=q / 4 \pi \varepsilon_{0} r\) and \(E=q / 4 \pi \varepsilon_{0} r^{2}\) giving E r=V \(2.0 \times 10^{6} \times 1.7 \times 10^{-2}=V\) \(V=3.4 \times 10^{4} V\)

Question 5

5

3 marks

Question 5(b)

5(b)

An isolated conducting sphere is charged. Fig. 5.1 shows the variation of the potential V due to the sphere with displacement x from its centre. Use Fig. 5.1 to determine:

structured3 marks

Question 5(b)(i)

5(b)(i)

the radius of the sphere radius = m

Mediumstructured1 marks

Answer

radius \(=0.060 \mathrm{~m}\) A1

Question 5(b)(ii)

5(b)(ii)

the charge on the sphere. charge = C

Mediumstructured2 marks

Answer

\[ \begin{aligned} & V=Q / 4 \pi \varepsilon_{0} x & Q=(-) 850 \times 4 \pi \times 8.85 \times 10^{-12} \times 0.060 & \text { or } & Q=(-) 850 \times 0.060 / 8.99 \times 10^{9} \end{aligned} \] (any correct pair of V and x values from curve) C1 \(Q=-5.7 \times 10^{-9} \mathrm{C}\) A1