*Question*

The diagram shows the electric field lines of a positively charged conducting sphere of radius

Points A and B are located on the same field line.

(a) Explain why the electric potential decreases from A to B. [2]

(b) Draw, on the axes, the variation of electric potential *V *with distance *r *from the centre of the sphere. [2]

(c) The concept of potential is also used in the context of gravitational fields. Suggest why scientists developed a common terminology to describe different types of fields. [1]

**▶️Answer/Explanation**

**Ans: **

**a.**

ALTERNATIVE 1

work done on moving a positive test charge in any outward direction is negative ✓

potential difference is proportional to this work «so V decreases from A to B» ✓

ALTERNATIVE 2

potential gradient is directed opposite to the field so inwards ✓

the gradient indicates the direction of increase of V «hence V increases towards the

centre/decreases from A to B» ✓

ALTERNATIVE 3

V = \(\frac{KQ}{R}\)

so as r increases V decreases ✓

V is positive as Q is positive ✓

ALTERNATIVE 4

the work done per unit charge in bringing a positive charge from infinity ✓

to point B is less than point A ✓

** b**

curve decreasing asymptotically for r > R

non – zero constant between 0 and R

** c** to highlight similarities between «different» fields

A planet has radius *R*. At a distance *h *above the surface of the planet the gravitational field strength is *g *and the gravitational potential is *V*.

a.i.State what is meant by gravitational field strength.[1]

*V*= –

*g*(

*R*+

*h*).[2]

*V*of the planet with height

*h*above the surface of the planet.

^{6}m. At a point P a distance 2.4 × 10

^{7}m above the surface of the planet the gravitational field strength is 2.2 N kg

^{–1}. Calculate the gravitational potential at point P, include an appropriate unit for your answer. [1]

When the asteroid was far away from the planet it had negligible speed. Estimate the speed of the asteroid at point P as defined in (b). [3]

^{12}kg. Calculate the gravitational force experienced by the

**planet**when the asteroid is at point P. [2]

**▶️Answer/Explanation**

## Markscheme

a.i.

the **«**gravitational**» **force per unit mass exerted on a point/small/test mass

**[1 mark]**

at height *h *potential is *V* = –\(\frac{{GM}}{{(R + h)}}\)

field is *g *= \(\frac{{GM}}{{{{(R + h)}^2}}}\)

**«**dividing gives answer**»**

*Do not allow an answer that starts with g = –*\(\frac{{\Delta V}}{{\Delta r}}\)* and then cancels the deltas and substitutes **R *+ *h*

*[2 marks]*

correct shape and sign

non-zero negative vertical intercept

**[2 marks]**

*V* = **«**–2.2 × (3.1 × 10^{6} + 2.4 × 10^{7}) =**»** **«**–**»** 6.0 × 10^{7} J kg^{–1}

*Unit is essential*

*Allow eg MJ kg ^{–}*

^{1}*if power of 10 is correct*

*Allow other correct SI units eg m*^{2}*s ^{–}*

^{2}

*, N m kg*

^{–}^{1}

*[1 mark]*

total energy at P = 0 / KE gained = GPE lost

**«**\(\frac{1}{2}\)*mv*^{2} + *mV* = 0 ⇒**»** *v* = \(\sqrt { – 2V} \)

*v* = **«**\(\sqrt {2 \times 6.0 \times {{10}^7}} \) =**»** 1.1 × 10^{4} **«**ms^{–1}**»**

*Award **[3] **for a bald correct answer*

*Ignore negative sign errors in the workings*

*Allow ECF from 6(b)*

*[3 marks]*

*ALTERNATIVE 1*

force on asteroid is **«**6.2 × 10^{12} × 2.2 =**»** 1.4 × 10^{13} **«**N**»**

**«**by Newton’s third law**» **this is also the force on the planet

*ALTERNATIVE 2*

mass of planet = 2.4 x 10^{25} **«**kg**» «**from *V* = –\(\frac{{GM}}{{(R + h)}}\)**»**

force on planet **«**\(\frac{{GMm}}{{{{(R + h)}^2}}}\)**»** = 1.4 × 10^{13} **«**N**»**

*MP2 must be explicit*

*[2 marks]*