· H2 Physics — Ten-Year Series, worked ← All 7 questions

2022 A-Level · Paper 2 · Question 3

The same parabola for a mass and a charge, and where two unlike charges cancel

Gravitational field · electric field 9 marks Uniform fields Field strength and potential Neutral point

Nine marks comparing gravitational and electric fields. A mass and a charge entering uniform fields bend along the same curve; the gravitational and electric forces between two protons have opposite signs and wildly different sizes. Then two unlike charges 6.0 cm apart: is the field zero between them, is the potential, and where would an electron feel no force at all? Below is the full solution, and beside it the path and the line through the charges.

The model, live

Path / field strength EPotential VForceZero

Part (a)

2 marks

An object enters a uniform field moving perpendicular to the field. Explain why the path has the same shape whether it is a mass in a gravitational field or a charge in an electric field.

Approach

The shape of the path depends only on the force, not on what causes it. In both cases a uniform field gives a force that is constant in size and direction, perpendicular to the initial velocity.

In both cases the field is uniform, so the force on the object is constant in magnitude and direction, and it acts perpendicular to the initial velocity. The object therefore has a constant acceleration along the field and a constant velocity perpendicular to it, which gives the same parabolic path.

Part (b)(i)

1 mark

The graphs show the gravitational force FGF_G and the electric force FEF_E between two protons against separation. Explain why they have opposite signs.

The gravitational force between the protons is attractive (negative), while the electric force between two like (positive) charges is repulsive (positive).

Part (b)(ii)

2 marks

State and explain whether the axes on both graphs are drawn to the same scale.

Approach

The curves look about the same height. Compare the actual sizes: both forces go as 1/r21/r^2, so their ratio is the same at every separation.

FEFG=e2/4πε0Gmp2=(9.0×109)(1.60×10−19)2(6.67×10−11)(1.67×10−27)2=1.2×1036\frac{F_E}{F_G} = \frac{e^2/4\pi\varepsilon_0}{Gm_p^2} = \frac{(9.0\times10^{9})(1.60\times10^{-19})^2}{(6.67\times10^{-11})(1.67\times10^{-27})^2} = 1.2\times10^{36}

No. The electric force between two protons is about 103610^{36} times the gravitational force at every separation, so if the curves look the same size the force axes must be on very different scales. (The separation axes could be the same.)

Part (c)(i)

1 mark

P (+6.5 nC+6.5\,\text{nC}) and R (−0.45 nC-0.45\,\text{nC}) are 6.0 cm6.0\,\text{cm} apart. Explain whether the electric field strength is zero anywhere on the line between them.

No. Between the charges the field due to P points away from P (towards R), and the field due to R points towards R. Both point the same way, so they add and cannot cancel.

Part (c)(ii)

1 mark

Explain whether the electric potential is zero anywhere on the line between them.

Yes. Potential is a scalar: P contributes a positive potential and R a negative one. Near P the total is positive and near R it is negative, so it passes through zero at a point between them (close to R, since R's charge is smaller).

Note

The contrast is vectors against scalars. Two fields add by direction, and between unlike charges the directions agree. Two potentials add by sign, and here the signs differ.

Part (c)(iii)

2 marks

An electron is on the line through P and R. Find its distance from P where there is no resultant force on it.

Approach

No force means zero field. Between the charges that is impossible (part (i)). Outside, the two fields point opposite ways, and they can only balance on the side of the smaller charge, R, where the nearer, weaker charge can match the further, stronger one.

At distance dd from P, beyond R:

6.5d2=0.45(d−6.0)2⟹dd−6.0=6.50.45=3.80\frac{6.5}{d^2} = \frac{0.45}{(d - 6.0)^2} \quad\Longrightarrow\quad \frac{d}{d - 6.0} = \sqrt{\frac{6.5}{0.45}} = 3.80
d=3.80d−22.8⟹d=22.82.80=8.1 cmd = 3.80d - 22.8 \quad\Longrightarrow\quad d = \frac{22.8}{2.80} = 8.1\,\text{cm}

To check your answer:

At 8.14 cm8.14\,\text{cm} from P the point is 2.14 cm2.14\,\text{cm} beyond R. 6.5/8.142=0.0986.5/8.14^2 = 0.098 and 0.45/2.142=0.0980.45/2.14^2 = 0.098. Equal.

distance =8.1 cm= 8.1\,\text{cm} from P (on the far side of R)

Watch

The other root of the quadratic, about 4.7 cm4.7\,\text{cm}, lies between the charges, where the fields add. It has to be rejected.

What this question was really testing

  1. Same force pattern, same motion. A uniform field gives a parabola whatever the field is.
  2. Attraction and repulsion as signs.
  3. Size. Electric forces between particles dwarf gravitational ones by about 103610^{36}.
  4. Vectors cancel by direction, scalars by sign. The field cannot vanish between unlike charges; the potential must.
  5. Where a neutral point lives. Outside the charges, nearer the smaller one.

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