Answer to Question #94952 in Electricity and Magnetism for abdul

Question

Two particles are fixed to an x axis: particle 1 of charge -7.40 × 10-7 C is at the origin and particle 2 of charge +7.40 × 10-7 C is at x2 = 19.1 cm. Midway between the particles, what is the magnitude of the net electric field?

Answer 1

The magnitude of the electric field, created at distance $r$ from the point charge $q$ is $\frac{k q}{r^2}$ , where $k = \frac{1}{4 \pi \varepsilon_0} = 8.99 \cdot 10^9 N m^2/ C^2$ - Coulomb's constant.

Since the charges are equal in absolute value, and the distance from the midpoint to them is the same, the magnitude of the electric field, created by each charge is equal. The direction of the field, created by the positive charge is West, same as one, created by the negative charge.

Hence, the magnitude of the net electric field is $|E| = 2 \frac{k q}{r^2}$ , where $q = 7.40 \cdot 10^{-7} C$ is the absolute value of both charges and $r = x_2 /2 = 9.55 cm$. Substituting the values, obtain $|E| =$ $139.32 \cdot 10^3 \frac{V}{m}$.