# IBDP Physics 10.1 – Describing fields: IB style Question Bank HL Paper 1

### Question

Which is a correct unit for gravitational potential?

A

B J kg

C m

D N

Ans: A

GRAVITATIONAL POTENTIAL

Gravitational potential at any point is the work done by gravitational force in carrying a body of unit mass from infinity to that point in gravitational influence of source. i.e.,
where r is the distance of point from source mass M.
The S.I. unit of V is joule/kg. It can be expressed in $$m^2s^{-2}$$. Its dimensions are [M0L2T–2]. It is a scalar quantity.

### Question

A positive point charge is placed above a metal plate at zero electric potential. Which diagram shows the pattern of electric field lines between the charge and the plate?

### Markscheme

C

Electric lines of force generate or terminate at charges /surfaces at right angles. Positive charge induce a negative charge towards it positive charge on other side of plate.

### Question

The diagram shows the electric field and the electric equipotential surfaces between two charged parallel plates. The potential difference between the plates is 200 V.

What is the work done, in nJ, by the electric field in moving a negative charge of magnitude 1 nC from the position shown to X and to Y?

### Markscheme

A

$$W_{AB} = Q(V_B – V_A)$$
As per given question
C to X  $$\Delta V= 0$$, hence $$U_{CX} = Q\Delta V =0$$ or dU=-dW =0
C to Y $$\Delta V= 50$$
work done by the electric field during the displacement is given by Change in potential energy
or

$$W_{CX} =-10^{-9}\times 50 = -50 \; nJ$$

dU=-dW
or
$$dW = -(-50) = 50 \; nJ$$

### Question

A moon of mass M orbits a planet of mass 100M. The radius of the planet is R and the distance between the centres of the planet and moon is 22R.

What is the distance from the centre of the planet at which the total gravitational potential has a maximum value?

A. 2R

B. 11R

C. 20R

D. 2R and 20R

### Markscheme

C

Let the potential be maximum at point x from center of planet

Gravitational potential is given by

$$V_g=-\frac{G100M}{x}-\frac{GM}{22-x}$$
now  $$V_g$$  to be maximum  $$\frac{dV_g}{dx}=0$$
or
$$\frac{d}{dx}(-\frac{G100M}{x}-\frac{GM}{22-x})=0$$
or
$$\frac{d}{dx}(-\frac{100}{x}-\frac{1}{22-x})=0$$
or
$$\frac{100}{x^2}-\frac{1}{(22-x)^{2}}=0$$
$$100 \times (22-x)^2=x^2$$
or
$$10 \times (22-x) =x$$
$$220 – 10x =x$$
$$11 x =220$$
$$x = 20$$

### Question

The diagram shows 5 gravitational equipotential lines. The gravitational potential on each line is indicated. A point mass m is placed on the middle line and is then released. Values given in MJ kg–1.

Which is correct about the direction of motion and the acceleration of the point mass?

### Markscheme

D

we have relation in force and potential as

$$F= -\frac{dU}{dx}=-\frac{\Delta U}{\Delta x}=\frac{U_1-U_2}{x_2-x_1}$$

Hence we can see gravitational force will be towards left and this will cause acceleration of the point mass

### Question

Four identical, positive, point charges of magnitude Q are placed at the vertices of a square of side 2d. What is the electric potential produced at the centre of the square by the four charges?

A. 0

B. $$\frac{{4kQ}}{d}$$

C. $$\frac{{\sqrt 2 kQ}}{d}$$

D. $$\frac{{2\sqrt 2 kQ}}{d}$$

### Markscheme

D

All four charges are equidistance  from center square.

Hence distance for each charge $$= \frac{\sqrt{(2d)^2+(2d)^2}}{2}=\sqrt{2}d$$
potential due to a charge Q at distance $$\sqrt{2}d$$  is
$$V_{p1} =k\frac{Q}{\sqrt{2}d}$$
Hence due to all four charge
$$4 \times V_{p1}$$  all four charges are same and equidistance from center
$$\therefore V_p = 4 \times V_{p1} = 4 \times k\frac{Q}{\sqrt{2}d}$$
$$=\frac{2\sqrt{2} kQ}{d}$$