*Question*

Which is a correct unit for gravitational potential?

A

B J kg

C m

D N

**▶️Answer/Explanation**

Ans: A

GRAVITATIONAL POTENTIAL

*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?

**▶️Answer/Explanation**

### 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?

**▶️Answer/Explanation**

### Markscheme

A

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

dU=-dW

or

\(dW = -(-50) = 50 \; nJ\)

*Question*

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

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

A. 2*R*

B. 11*R*

C. 20*R*

D. 2*R *and 20*R*

**▶️Answer/Explanation**

### 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?

**▶️Answer/Explanation**

### 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 2*d*. 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. \(\frac{{2\sqrt 2 kQ}}{d}\)

**▶️Answer/Explanation**

### 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

\(\therefore V_p = 4 \times V_{p1} = 4 \times k\frac{Q}{\sqrt{2}d}\)

\(=\frac{2\sqrt{2} kQ}{d}\)