Worked examples · Sheet 2

Work done by the electric force and potential differences

Theory used: Work done by the electric force Electric potential Potential difference and field Potential difference in a uniform field Circulation of the electrostatic field Work in the field of a point charge Potential of a point charge Electronvolt

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

In a region of space there is a uniform electric field $E = E \overset{}{^}$ with $E = 2.5 \cdot 1 0^{3} V/m$ . A charge $q = - 4.0 μ C$ moves from the point $A = (0, 0)$ to the point $B = (0.40 m, 0.30 m)$ . Compute the potential difference $V_{B} - V_{A}$ , the work done by the electric force and the work done by an external agent that moves the charge slowly.

In the uniform field directed along

x

the work from

A

B

is the same along the segment

A B

and along the broken line

A C B

, where only the segment

A C

, parallel to the field, contributes

V_{B} - V_{A} = - E \cdot A B = - E Δ x, W_{A \to B} = - q (V_{B} - V_{A}), W_{ext} = - W_{A \to B}

V_{B} - V_{A} W_{A \to B} W_{ext} = - 2.5 \cdot 1 0^{3} \cdot 0.40 V = - 1.0 \cdot 1 0^{3} V = - (- 4.0 \cdot 1 0^{- 6}) \cdot (- 1.0 \cdot 1 0^{3}) J = - 4.0 \cdot 1 0^{- 3} J = + 4.0 \cdot 1 0^{- 3} J

The field is uniform, so the potential difference in a uniform field applies: only the projection of the displacement onto the direction of the field matters, which is $Δ x = 0.40 m$ , while the displacement $Δ y = 0.30 m$ is perpendicular to the field and does not change the potential. The potential decreases in the direction of the field, therefore $V_{B} < V_{A}$ . The work done by the field is $- q Δ V$ ; the charge is negative and moves toward lower potentials, so its potential energy increases and the work done by the field is negative. The same result is obtained along the broken line $A C B$ : on the segment $A C$ the force $q E$ is opposite to the displacement and does the work $q E \cdot 0.40 m = - 4.0 \cdot 1 0^{- 3} J$ , on the segment $C B$ it is perpendicular and does no work, as predicted by the zero circulation of the electrostatic field. If the charge moves slowly, the kinetic energy does not change and the total work is zero, so the external agent does the work opposite to that of the field.

Exercise 2

A point charge $Q = 5.0 nC$ is fixed. A charge $q = - 2.0 nC$ is brought slowly from the point $A$ , at distance $r_{A} = 10 cm$ from $Q$ , to the point $B$ , at distance $r_{B} = 25 cm$ . Compute the potentials at $A$ and at $B$ , the work done by the electric force and the external work.

V_{A} = \frac{k Q}{r _{A}}, V_{B} = \frac{k Q}{r _{B}}, W_{A \to B} = q (V_{A} - V_{B}) = k Qq (\frac{1}{r _{A}} - \frac{1}{r _{B}}), W_{ext} = - W_{A \to B}

V_{A} V_{B} W_{A \to B} W_{ext} = \frac{8.99 \cdot 1 0 ^{9} \cdot 5.0 \cdot 1 0 ^{- 9}}{0.10} V \approx 450 V = \frac{8.99 \cdot 1 0 ^{9} \cdot 5.0 \cdot 1 0 ^{- 9}}{0.25} V \approx 180 V = - 2.0 \cdot 1 0^{- 9} \cdot (449.5 - 179.8) J \approx - 5.4 \cdot 1 0^{- 7} J \approx + 5.4 \cdot 1 0^{- 7} J

The charge $Q$ is pointlike, so the potentials are computed with the potential of a point charge, with the reference at infinity. The work done by the field depends only on the initial and final potentials, as guaranteed by the work in the field of a point charge, and the path followed is not needed. The charges have opposite signs, so they attract each other; when $q$ is moved away from $Q$ the electric force is opposite to the radial displacement and does negative work. Equivalently, the negative charge moves toward lower potentials and its potential energy $q V$ increases. The displacement is slow, therefore the external agent supplies the positive work $5.4 \cdot 1 0^{- 7} J$ , which becomes potential energy of the system.

Exercise 3

In a region of space the electric field is directed along the $x$ axis and has component $E_{x} (x) = b x^{2}$ , with $b = 6.0 \cdot 1 0^{3} V/ m^{3}$ . Compute the potential difference $V (2.0 m) - V (0)$ , the work done by the field on an electron that goes from $x = 0$ to $x = 2.0 m$ and the minimum kinetic energy, in electronvolts, that the electron must have at $x = 0$ to reach $x = 2.0 m$ .

V (x_{B}) - V (x_{A}) = - \int_{x_{A}}^{x_{B}} E_{x} (x) d x = - \frac{b}{3} (x_{B}^{3} - x_{A}^{3})

W = - q Δ V = e Δ V, K_{m i n} = - W = - e Δ V

Δ V W K_{m i n} = - \frac{6.0 \cdot 1 0 ^{3}}{3} \cdot 2. 0^{3} V = - 1.6 \cdot 1 0^{4} V = 1.60 \cdot 1 0^{- 19} \cdot (- 1.6 \cdot 1 0^{4}) J \approx - 2.6 \cdot 1 0^{- 15} J = 1.6 \cdot 1 0^{4} eV = 16 keV

The field varies with position, so the potential difference is computed with the integral of the potential difference and field. One integrates along the $x$ axis, where $d ℓ = d x \overset{}{^}$ and $E \cdot d ℓ = E_{x} d x$ ; an antiderivative of $b x^{2}$ is $\frac{b x ^{3}}{3}$ . The field points toward increasing $x$ , therefore the potential decreases from $x = 0$ to $x = 2.0 m$ . The electron has charge $q = - e$ and moves toward lower potentials, so its potential energy increases and the field does negative work, that is, it slows the electron down. By the work-energy theorem the electron reaches $x = 2.0 m$ only if its initial kinetic energy is at least equal to the resistive work, $16 keV$ ; in electronvolts the result is read directly from $Δ V$ , because the charge is $e$ .