Question bank
Practice A-Level CAIE Physics 18 5 Electric Potential questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.
Question 1
Question 1(c)
State one similarity and one difference between gravitational potential due to a point mass and electric potential due to a point charge. similarity difference
similarity - any one point from - inversely proportional to distance (from point) - points of equal potential lie on concentric spheres - zero at infinite distance B1 difference - any one point from - gravitational potential is (always) negative - electric potential can be positive or negative B1
Question 3
Question 3(a)
Define electric potential at a point.
work done bringing unit positive charge from infinity (to the point)
Question 3(b)
Two point charges A and B are separated by a distance of 20 nm in a vacuum, as illustrated in Fig. 3.1. A point P is a distance x from A along the line A B. The variation with distance x of the electric potential \(V_{A}\) due to charge A alone is shown in Fig. 3.2. The variation with distance x of the electric potential \(V_{B}\) due to charge B alone is also shown in Fig. 3.2.
Question 3(b)(ii)
By reference to Fig. 3.2, state how the combined electric potential due to both charges may be determined.
the individual potentials are summed
Question 3(b)(iii)
Without any calculation, use Fig. 3.2 to estimate the distance x at which the combined electric potential of the two charges is a minimum.
allow value of x between 10 nm and 13 nm
Question 3(b)(iv)
The point P is a distance \(x=10 \mathrm{~nm}\) from A. An \(\alpha\)-particle has kinetic energy \(E_{\mathrm{K}}\) when at infinity. Use Fig. 3.2 to determine the minimum value of \(E_{\mathrm{K}}\) such that the \(\alpha\)-particle may travel from infinity to point P .
\(V=0.43 \mathrm{~V}\) (allow \(0.42 \mathrm{~V} \rightarrow 0.44 \mathrm{~V}\) ) energy \(=2 \times 1.6 \times 10^{-19} \times 0.43\)
Question 4
Question 4(a)
Define electric potential at a point.
work done in bringing unit positive charge from infinity (to that point)
Question 4(b)
Two small spherical charged particles P and Q may be assumed to be point charges located at their centres. The particles are in a vacuum. Particle P is fixed in position. Particle Q is moved along the line joining the two charges, as illustrated in Fig. 4.1. The variation with separation x of the electric potential energy \(E_{\mathrm{p}}\) of particle Q is shown in Fig. 4.2.
Question 4(b)(i)
State how the magnitude of the electric field strength is related to potential gradient.
field strength is potential gradient
Question 4(b)(ii)
Use your answer in (i) to show that the force on particle Q is proportional to the gradient of the curve of Fig. 4.2.
field strength proportional to force (on particle Q) potential gradient proportional to gradient of (potential energy) graph
Question 4(c)
The magnitude of the charge on each of the particles P and Q is \(1.6 \times 10^{-19} \mathrm{C}\). Calculate the separation of the particles at the point where particle Q has electric potential energy equal to -5.1 eV . separation = m
energy \(=5.1 \times 1.6 \times 10^{-19}(\mathrm{~J}) \quad \mathrm{C} 1\) potential energy \(=Q_{1} Q_{2} / 4 \pi \varepsilon_{0} r\) C1 \(5.1 \times 1.6 \times 10^{-19}=\left(1.6 \times 10^{-19}\right)^{2} / 4 \pi \times 8.85 \times 10^{-12} \times r \quad\) C1 \(r=2.8 \times 10^{-10} \mathrm{~m}\) A1
Question 4
Question 4(b)
A uniform electric field is produced by applying a potential difference of 1200 V across two parallel metal plates in a vacuum, as shown in Fig. 4.1. The separation of the plates is 14 mm . A particle P with charge \(3.2 \times 10^{-19} \mathrm{C}\) and mass \(6.6 \times 10^{-27} \mathrm{~kg}\) starts from rest at the lower plate and is moved vertically to the top plate by the electric field. Calculate
Question 4(b)(ii)
the work done on P by the electric field, work done = J
W=Q V or \(W=F \times d\) and therefore \(W=E \times Q \times d\) \(=3.2 \times 10^{-19} \times 1200\) \(=3.84 \times 10^{-16} \mathrm{~J}\)
Question 18
A positive charge experiences a force F when placed at point X in a uniform electric field. The charge is then moved from point X to point Y . Distances r and s are shown on the diagram. What is the change in the potential energy of the charge? decreases by Fs increases by Fs decreases by Fr increases by Fr
A
Question 4
Question 4(c)
The electric potential gradient is related to the electric field. Use data from Fig. 4.2 to state the value of x at which the magnitude of the electric potential gradient is maximum. Give a reason for the value you have chosen.
field strength or E is potential gradient or field strength is rate of change of (electric) potential M1 (field strength) maximum at \(x=6 \mathrm{~cm}\) A1
Question 5
An isolated solid metal sphere of radius r is given a positive charge. The distance from the centre of the sphere is x.
Question 5(a)
The electric potential at the surface of the sphere is \(V_{0}\). On the axes of Fig.5.1, sketch a graph to show the variation with distance x of the electric potential due to the charged sphere, for values of x from x=0 to x=4 r.
graph: straight line at constant potential \(=V_{0}\) from x=0 to \(x=r \quad\) B1 curve with decreasing gradient M1 passing through \(\left(2 r, 0.50 V_{0}\right)\) and \(\left(4 r, 0.25 V_{0}\right)\) A1
Question 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.
Question 4(b)
Explain why, without any calculation, when a small test charge is moved from point P to point Q, the net work done is zero.
potential at P is same as potential at \(\mathrm{Q} \quad\) B1 work done \(=q \Delta V \quad\) M1 \(\Delta V=0\) so zero work done A0
Question 4(c)
Calculate the work done by an electron in moving from the midpoint of line AB to point P.
at midpoint, potential is \(2 \times\left(6.4 \times 10^{-19}\right) /\left(4 \pi \varepsilon_{0} \times 6 \times 10^{-6}\right) \quad\) C1 at P , potential is \(\left(6.4 \times 10^{-19}\right) /\left(4 \pi \varepsilon_{0} \times 3 \times 10^{-6}\right)+\left(6.4 \times 10^{-19}\right) /\left(4 \pi \varepsilon_{0} \times 9 \times 10^{-6}\right) \quad \mathrm{C} 1\) change in potential \(=\left(6.4 \times 10^{-19}\right) /\left(4 \pi \varepsilon_{0} \times 9 \times 10^{-6}\right)\) energy \(=1.6 \times 10^{-19} \times\left(6.4 \times 10^{-19}\right) /\left(4 \pi \varepsilon_{0} \times 9 \times 10^{-6}\right) \quad\) C1
Question 4
Question 4(a)
Define electric potential at a point.
work done moving unit positive charge from infinity (to the point)
Question 4(b)
A charged particle is accelerated from rest in a vacuum through a potential difference V. Show that the final speed v of the particle is given by the expression where \(\frac{q}{m}\) is the ratio of the charge to the mass (the specific charge) of the particle.
(gain in) kinetic energy = change in potential energy \(\frac{1}{2} m v^{2}=q V\) leading to \(v=(2 \mathrm{~V} q / \mathrm{m})^{1 / 2}\)
Question 4(c)
A particle with specific charge \(+9.58 \times 10^{7} \mathrm{C} \mathrm{kg}^{-1}\) is moving in a vacuum towards a fixed metal sphere, as illustrated in Fig. 4.1. particle specific charge \(+9.58 \times 10^{7} \mathrm{C} \mathrm{kg}^{-1}\) metal sphere potential +470 V The initial speed of the particle is \(2.5 \times 10^{5} \mathrm{~ms}^{-1}\) when it is a long distance from the sphere. The sphere is positively charged and has a potential of +470 V . Use the expression in (b) to determine whether the particle will reach the surface of the sphere.
either \(\left(2.5 \times 10^{5}\right)^{2}=2 \times V \times 9.58 \times 10^{7}\) \(V=330 \mathrm{~V}\) this is less than 470 V and so 'no' A1 or \(\quad v=\left(2 \times 470 \times 9.58 \times 10^{7}\right)\) \(v=3.0 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}\) this is greater than \(2.5 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}\) and so 'no' or \(\quad\left(2.5 \times 10^{5}\right)^{2}=2 \times 470 \times(q / m)\) \((q / m)=6.6 \times 10^{7} \mathrm{Ckg}^{-1}\) this is less than \(9.58 \times 10^{7} \mathrm{Ckg}^{-1}\) and so 'no'
Question 5
Question 5(a)
Define electric potential at a point.
work done in moving unit positive charge M1 from infinity (to the point) A1 [2]