Worked examples · Sheet 2

Fields of point charges in the plane

Theory used: Field of a point charge Superposition principle for the electric field somma vettoriale dei campi Force on a charge placed in the field

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Two equal charges $q = 5.0 nC$ are at rest at the points $A = (- 4.0 cm, 0)$ and $B = (4.0 cm, 0)$ . Compute the electric field at the point $P = (0, 3.0 cm)$ .

r = 4. 0^{2} + 3. 0^{2} cm = 5.0 cm, E_{1} = E_{2} = k \frac{q}{r ^{2}}

E_{x} = - E_{1} \frac{4}{5} + E_{2} \frac{4}{5} = 0, E_{y} = 2 E_{1} \frac{3}{5}

E_{1} E_{y} = \frac{8.99 \cdot 1 0 ^{9} \cdot 5.0 \cdot 1 0 ^{- 9}}{0.05 0 ^{2}} N/C \approx 1.80 \cdot 1 0^{4} N/C = 2 \cdot 1.80 \cdot 1 0^{4} \cdot 0.60 N/C \approx 2.2 \cdot 1 0^{4} N/C

The point $P$ is equidistant from the two charges, so by the field of a point charge the two fields have the same magnitude. The field of $A$ points away from $A$ and has the direction of the unit vector $(\frac{4}{5}, \frac{3}{5})$ , which goes from $A$ to $P$ ; the field of $B$ has the direction of the unit vector $(- \frac{4}{5}, \frac{3}{5})$ . By the superposition principle the components are added: the $x$ components are opposite and cancel, while the $y$ components are equal and add up. The field at $P$ is therefore directed along the $y$ -axis, upward, and equals $2.2 \cdot 1 0^{4} N/C$ ; the result respects the symmetry of the figure with respect to the $y$ -axis.

Exercise 2

A charge $q_{1} = 3.0 nC$ is at the origin and a charge $q_{2} = - 6.0 nC$ is at the point $(0, 4.0 cm)$ . Compute the magnitude and direction of the electric field at the point $P = (3.0 cm, 0)$ .

P

the field

E_{1}

points away from

q_{1}

, the field

E_{2}

is directed toward

q_{2}

; the total field

E

is the diagonal of the parallelogram

E_{1} = k \frac{q _{1}}{r _{1}^{2}}, E_{2} = k \frac{∣ q _{2} ∣}{r _{2}^{2}}, r_{1} = 3.0 cm, r_{2} = 5.0 cm

E_{1} = E_{1} \overset{}{^}, E_{2} = E_{2} (- \frac{3}{5} \overset{}{^} + \frac{4}{5} \overset{}{^})

E_{1} E_{2} E_{x} E_{y} E = \frac{8.99 \cdot 1 0 ^{9} \cdot 3.0 \cdot 1 0 ^{- 9}}{0.03 0 ^{2}} N/C \approx 3.00 \cdot 1 0^{4} N/C = \frac{8.99 \cdot 1 0 ^{9} \cdot 6.0 \cdot 1 0 ^{- 9}}{0.05 0 ^{2}} N/C \approx 2.16 \cdot 1 0^{4} N/C = (3.00 - 0.60 \cdot 2.16) \cdot 1 0^{4} N/C \approx 1.70 \cdot 1 0^{4} N/C = 0.80 \cdot 2.16 \cdot 1 0^{4} N/C \approx 1.73 \cdot 1 0^{4} N/C = E_{x}^{2} + E_{y}^{2} \approx 2.4 \cdot 1 0^{4} N/C, φ = arctan \frac{E _{y}}{E _{x}} \approx 4 5^{\circ}

The two charges are at different distances from $P$ : $q_{1}$ is $3.0 cm$ away and $q_{2}$ is $5.0 cm$ away, by the Pythagorean theorem in the right triangle with legs $3.0 cm$ and $4.0 cm$ . The field of $q_{1}$ , positive, points outward, so at $P$ it is directed along $\overset{}{^}$ . The field of $q_{2}$ , negative, is directed from $P$ toward $q_{2}$ , that is along the unit vector $(- \frac{3}{5}, \frac{4}{5})$ .

By the superposition principle the components are added. Along $x$ the contribution of $q_{2}$ is negative and reduces that of $q_{1}$ ; along $y$ only $q_{2}$ contributes. The resultant field has magnitude $2.4 \cdot 1 0^{4} N/C$ and forms an angle of about $4 5^{\circ}$ with the $x$ -axis, in the first quadrant. The magnitude is smaller than the sum of the magnitudes, $5.2 \cdot 1 0^{4} N/C$ , because the two fields form an obtuse angle.

Exercise 3

Three equal charges $q = 2.0 nC$ occupy three vertices of a square of side $a = 10 cm$ . Compute the electric field at the fourth vertex $D$ and the force that a charge $q_{0} = - 1.0 nC$ placed at $D$ would experience.

E_{side} = k \frac{q}{a ^{2}}, E_{diag} = k \frac{q}{( a 2 ) ^{2}} = \frac{1}{2} k \frac{q}{a ^{2}}

E = 2 E_{side} cos 4 5^{\circ} + E_{diag} = k \frac{q}{a ^{2}} (2 + \frac{1}{2})

k \frac{q}{a ^{2}} E F = \frac{8.99 \cdot 1 0 ^{9} \cdot 2.0 \cdot 1 0 ^{- 9}}{0.1 0 ^{2}} N/C \approx 1.80 \cdot 1 0^{3} N/C \approx 1.80 \cdot 1 0^{3} \cdot 1.914 N/C \approx 3.4 \cdot 1 0^{3} N/C = ∣ q_{0} ∣ E \approx 1.0 \cdot 1 0^{- 9} \cdot 3.44 \cdot 1 0^{3} N \approx 3.4 \cdot 1 0^{- 6} N

Choose a reference frame with $D$ at the origin and the sides of the square along the axes, so that the charges are at $(- a, 0)$ , $(0, - a)$ and $(- a, - a)$ . The two charges at the vertices adjacent to $D$ are at distance $a$ and produce fields of magnitude $k \frac{q}{a ^{2}}$ , pointing outward, directed along $\overset{}{^}$ and along $\overset{}{^}$ . The charge at the opposite vertex is at a distance equal to the diagonal, $a 2$ , and produces a field of half the magnitude, directed along the diagonal, outward from the square.

By symmetry with respect to the diagonal the total field is directed along the diagonal. Each of the two fields of the adjacent vertices forms an angle of $4 5^{\circ}$ with the diagonal and contributes the projection $k \frac{q}{a ^{2}} cos 4 5^{\circ}$ , while the third field is already directed along the diagonal. The field at $D$ is therefore $3.4 \cdot 1 0^{3} N/C$ and is directed along the diagonal, outward. The charge $q_{0}$ is negative, therefore the force has orientation opposite to the field, that is it is directed toward the center of the square, and has magnitude $3.4 \cdot 1 0^{- 6} N$ .