Worked examples · Sheet 1

Fields of collinear point charges

Theory used: Electric field Force on a charge placed in the field Field of a point charge Superposition principle for the electric field

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A point charge $Q = 3.0 nC$ is at rest in vacuum. Compute the electric field at the point $P$ at distance $r = 15 cm$ from $Q$ and the force experienced by an electron placed at $P$ .

E = k \frac{Q}{r ^{2}}, F = e E

E F = \frac{8.99 \cdot 1 0 ^{9} \cdot 3.0 \cdot 1 0 ^{- 9}}{0.1 5 ^{2}} N/C = \frac{26.97}{0.0225} N/C \approx 1.2 \cdot 1 0^{3} N/C = 1.60 \cdot 1 0^{- 19} \cdot 1.20 \cdot 1 0^{3} N \approx 1.9 \cdot 1 0^{- 16} N

The charge is a point charge at rest, so the formula for the field of a point charge applies. Since $Q > 0$ , the field at $P$ is directed along the joining line, pointing away from $Q$ . The force on a charge placed in the field is $F = q E$ ; the electron has charge $- e$ , therefore the force has magnitude $e E$ and orientation opposite to the field, that is the electron is attracted toward $Q$ , as Coulomb’s law predicts for charges of opposite sign. The results have two significant figures, like the data.

Exercise 2

Two point charges $q_{1} = 2.0 nC$ and $q_{2} = 8.0 nC$ are at rest on the $x$ -axis, at $x = 0$ and at $x = d = 30 cm$ . Compute the field at the midpoint $M$ of the segment joining them and find the point of the $x$ -axis at which the field is zero.

E_{x} (M) = k \frac{q _{1}}{( \frac{d}{2} ) ^{2}} - k \frac{q _{2}}{( \frac{d}{2} ) ^{2}} = \frac{4 k ( q _{1} - q _{2} )}{d ^{2}}

E_{x} (M) = \frac{4 \cdot 8.99 \cdot 1 0 ^{9} \cdot ( 2.0 - 8.0 ) \cdot 1 0 ^{- 9}}{0.3 0 ^{2}} N/C \approx - 2.4 \cdot 1 0^{3} N/C

k \frac{q _{1}}{x ^{2}} = k \frac{q _{2}}{( d - x ) ^{2}} ⟹ \frac{d - x}{x} = \frac{q _{2}}{q _{1}} = 2 ⟹ x = \frac{d}{3} = 10 cm

By the superposition principle the field at $M$ is the sum of the fields of the two charges, both directed along the $x$ -axis. The field of $q_{1}$ , positive, points away from $q_{1}$ , so at $M$ it points toward increasing $x$ ; the field of $q_{2}$ points away from $q_{2}$ , so at $M$ it points toward decreasing $x$ . The distances are equal and $q_{2}$ is larger, therefore the field at $M$ has magnitude $2.4 \cdot 1 0^{3} N/C$ and is directed toward $q_{1}$ .

The field can vanish only where the two fields have opposite orientations. Outside the segment the two positive charges produce fields with the same orientation, so the point sought lies between the charges, at distance $x$ from $q_{1}$ and $d - x$ from $q_{2}$ . Equating the magnitudes and taking the square root, with $0 < x < d$ , we find $x = 10 cm$ : the zero-field point is closer to the smaller charge, as it must be because the field decreases with distance.

Exercise 3

A charge $q_{1} = 4.0 nC$ is at $x = 0$ and a charge $q_{2} = - 1.0 nC$ is at $x = d = 20 cm$ . Compute the field at the point $x = 10 cm$ and find the point of the $x$ -axis, at finite distance, at which the field is zero.

E_{x} (10 cm) = k \frac{q _{1}}{0.1 0 ^{2}} + k \frac{∣ q _{2} ∣}{0.1 0 ^{2}} = \frac{8.99 \cdot 1 0 ^{9} \cdot 5.0 \cdot 1 0 ^{- 9}}{0.010} N/C \approx 4.5 \cdot 1 0^{3} N/C

x > d : k \frac{q _{1}}{x ^{2}} = k \frac{∣ q _{2} ∣}{( x - d ) ^{2}} ⟹ \frac{x}{x - d} = \frac{q _{1}}{∣ q _{2} ∣} = 2 ⟹ x = 2 d = 40 cm

At the point

P

40 cm

from

q_{1}

and

20 cm

from

q_{2}

, the fields

E_{1}

and

E_{2}

have the same magnitude and opposite orientations

At $x = 10 cm$ the field of $q_{1}$ , positive, points outward and toward increasing $x$ ; the field of $q_{2}$ , negative, is directed toward $q_{2}$ , that is also toward increasing $x$ . Since the two fields have the same orientation, by the superposition principle the magnitudes add up and the field is $4.5 \cdot 1 0^{3} N/C$ , in the direction of increasing $x$ .

To find the zero-field point we examine the three regions of the axis. Between the charges the fields have the same orientation, as just seen, so the sum does not vanish. For $x < 0$ the fields have opposite orientations, but the point is closer to $q_{1}$ , which has the larger magnitude, therefore the field of $q_{1}$ prevails. There remains the region $x > d$ , where the fields are opposite and the point is closer to the charge of smaller magnitude; equating the magnitudes gives $x = 40 cm$ . The equation between the magnitudes written without taking the regions into account also has the solution $x \approx 13 cm$ , which falls between the charges, where the fields have the same orientation, and therefore must be discarded. The field is zero only at $x = 40 cm$ , where both fields equal $2.2 \cdot 1 0^{2} N/C$ .