Exercise practice: Electric Charges and Fields

6.1 Exercise 1.1: force between two charged spheres

Two small charged spheres carry 2 × 10^-7 C and 3 × 10^-7 C and are placed 30 cm apart in air. The product of the charges is 6 × 10^-14 C^2 and the square of the distance is 0.09 m^2. Using k = 9 × 10^9, the force is 9 × 10^9 × 6 × 10^-14 / 0.09, which works out to 6 × 10^-3 N. The force is repulsive because both charges are positive.

6.2 Exercise 1.2: distance from a known force

A force of 0.2 N acts between a charge of 0.4 microcoulomb and a charge of -0.8 microcoulomb. Rearrange Coulomb's law to solve for r squared: r^2 = k q1 q2 / F. Substitute 9 × 10^9, the product 3.2 × 10^-13 and the force 0.2. The square is 1.44 × 10^-2 m^2, so the distance is 0.12 m. The second charge feels 0.2 N, attractive this time, because the two charges have opposite signs.

6.3 Exercise 1.6: forces at the centre of a square

Four charges of 2, -5, 2 and -5 microcoulomb occupy the corners of a square of side 10 cm in that order. A charge of 1 microcoulomb at the centre feels four forces. The pair on the diagonal AC is equal and opposite, and the pair on the diagonal BD is equal and opposite, so the total force is zero. Write the symmetry argument and the answer follows without any trigonometry.

6.4 Exercise 1.8: field at the midpoint

Charges of 3 microcoulomb and -3 microcoulomb are 20 cm apart. At the midpoint each charge is 0.1 m away. The two fields have equal size k q / r^2, which is 9 × 10^9 × 3 × 10^-6 / 0.01, so 2.7 × 10^6 N C^-1 each. Both fields point toward the negative charge, so they add to 5.4 × 10^6 N C^-1. A test charge of -1.5 × 10^-9 C placed there feels a force of 8.1 × 10^-3 N. The field pushes a negative charge opposite to its own direction, so this force points toward the positive charge.

6.5 Exercise 1.9: total charge and dipole moment

Two charges of 2.5 × 10^-7 C and -2.5 × 10^-7 C sit at (0, 0, -15 cm) and (0, 0, 15 cm). The total charge of the pair is zero, and the dipole moment is the charge times the separation. The direction runs from the negative charge to the positive one, which here means from +15 cm to -15 cm, so p = 2.5 × 10^-7 × 0.3 = 7.5 × 10^-8 C m along the negative z axis.

6.6 Exercise 1.10: torque on a dipole

A dipole of moment 4 × 10^-9 C m is at 30 degrees to a field of 5 × 10^4 N C^-1. The torque is p E sin 30, which is 4 × 10^-9 × 5 × 10^4 × 0.5 = 1 × 10^-4 N m. The torque vector is perpendicular to the plane that holds the dipole moment and the field.

6.7 Exercise 1.18: flux through a cube

A point charge of 2 microcoulomb is at the centre of a cubic surface of edge 9 cm. The flux equals the enclosed charge divided by epsilon0, so 2 × 10^-6 / 8.854 × 10^-12, which is 2.26 × 10^5 N m^2 C^-1. The edge length is not used at all, and that is deliberate: the question tests whether you know the flux rule.

6.8 Exercise 1.24 and the cone problem

A right circular cone of base radius R and height h sits on a table and a horizontal uniform field of size E passes through it. The flux entering the cone equals the flux leaving, because no charge is enclosed. The value asked for follows from the area projected on the field direction, and the sign depends on whether the question asks for entering or leaving flux. When the flux through a closed body is asked, split the surface into parts and use the dot product on each part.

6.9 Exercise 1.13: flux and charge

A point charge of 6 microcoulomb sits inside a closed surface. The net flux is q / epsilon0 = 6 × 10^-6 / 8.854 × 10^-12, about 6.8 × 10^5 N m^2 C^-1. If the charge is moved to another point inside the same surface, the flux does not change because it depends only on the enclosed charge. If the charge is moved outside the surface, the net flux becomes zero.