Consider the points A, B, and C on two equipotential surfaces as shown in the figure. When a proton moves from A to B, the electric field does a work of 3.2 × 10⁻¹⁹ J on it. Charge of an electron is -1.6 × 10⁻¹⁹ C. The electric potential differences V_(AB), V_(BC), and V_(CA), respectively, are

2019 A/L Physics Past Paper · Question 21 (English Medium)

Equipotential surfaces — potential differences

Consider the points $A$ , $B$ , and $C$ on two equipotential surfaces as shown in the figure. When a proton moves from $A$ to $B$ , the electric field does a work of $3.2 \times 10^{-19}\ \text{J}$ on it. Charge of an electron is $-1.6 \times 10^{-19}\ \text{C}$ . The electric potential differences $V_{AB}$ , $V_{BC}$ , and $V_{CA}$ , respectively, are

(1) 2 V, −2 V, and 0 V
(2) 2 V, −2 V, and 2 V
(3) −2 V, 2 V, and 0 V
(4) 0.5 V, −0.5 V, and 0 V
(5) −0.5 V, 0.5 V, and 0 V

Answer and explanation

Answer: (1) 2 V, −2 V, and 0 V

Given information

We know:

A and C are on the same equipotential surface.

B is on the other surface.

From A to B, the field does work on the proton: $W_{AB} = 3.2 \times 10^{-19}$ J

The charge of an electron is $-1.6 \times 10^{-19}$ C.

A proton has the same size of charge, but positive: $q = +1.6 \times 10^{-19}$ C

What the question asks

We want: $V_{AB}$ , $V_{BC}$ and $V_{CA}$ .

Relevant theory

The potential difference between two points in an electric field is the work done in taking a unit positive charge from one point to the other (This work is independent of the path along which the charge is carried).

Since an electric field does not exist along an equipotential surface no work is needed to carry a charge along such a surface.

The work the field does on a charge $q$ from $A$ to $B$ is:

$W_{AB} = q\,(V_A - V_B)$

So the potential difference is the work for each unit of charge:

$V_{AB} = \frac{W_{AB}}{q}$

Required equations

$W_{AB} = q\,(V_A - V_B)$

$V_{AB} = \frac{W_{AB}}{q}$

Applying it to this question

In SI units:

$W_{AB} = 3.2 \times 10^{-19}$ J

$q = +1.6 \times 10^{-19}$ C

Both are already in SI units.

Reasoning and calculation

Start with A to B.

The field does positive work on the positive proton.

So A is at a higher potential than B.

$V_{AB} = \frac{W_{AB}}{q}$

$2$

$V_{AB} = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}}$

$-2$

$V_{BC} = V_B - V_A = -V_{AB}$

$0$

$V_{CA} = V_C - V_A$

Now match it to the options.

We found $2$ V, $-2$ V and $0$ V. That is option (1).

(2) has $V_{CA} = 2$ V. But C and A are on the same surface.

(3) has every answer turned round. It uses the electron's charge.

(4) and (5) divide the charge by the work. That is the wrong way up.

So only (1) fits.

Checks

$W / q$

Check the unit.

$\frac{\text{J}}{\text{C}} = \text{V}$

That is the unit of potential difference. Good.

A quick check.

Go round A → B → C → A and come back to the start.

The potential must come back to where it began.

$2 + (-2) + 0 = 0$

Good.

Conclusion

$V_{AB} = 2\ \text{V},\ V_{BC} = -2\ \text{V},\ V_{CA} = 0$

$V_{AB} = 2$ V.

C is at the same potential as A. So $V_{BC} = -2$ V.

C and A are on the same surface. So $V_{CA} = 0$ .

So the answer is option (1).

Using the electron's charge, $-1.6 \times 10^{-19}$ C, for the proton.

The proton's charge is positive: $+1.6 \times 10^{-19}$ C.

With the wrong charge, every answer turns round. That gives option (3).

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