Question

Find the magnitude of the electric field at a point midway between two charges +35.5 × 10−9 C and +77.2 × 10−9 C separated by a distance of 46.1 cm. The value of the Coulomb constant is 8.99 × 109 N · m2 /C 2 . Answer in units of N/C.

Answers

  1. The electric field at the midway between two charges is 7059.9N/C

    Explanation:

    Given-

    charge, q1 = 35.5 X 10⁻⁹ C

    q2 = 77.2 X 10⁻⁹C

    distance, r = 46.1cm

    midway distance, r = 23.05 cm = 0.2305 m

    Coulomb’s constant, k = 8.99 X 10⁹ Nm²/C²

    Electric field, E = ?

    We know,

    E = \frac{kq}{r^2} \\\\E = \frac{kq_2}{r^2} - \frac{kq_1}{r^2} \\\\E = \frac{k}{r^2} ( q_2 - q_1)\\\\

    E = \frac{8.99 X 10^9}{(0.2305)^2} (77.2 X 10^-^9 - 35.5 X10^-^9)\\\\E = \frac{8.99}{0.0531} (41.7)\\ \\E = 7059.9N/C

    Therefore, electric field at the midway between two charges is 7059.9N/C

Leave a Comment