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3 years ago

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Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷

C at x = 21.0 cm.
Midway between the particles, what is their net electric field?

1 answer:

victus00 [196]3 years ago

7 0

Answer:

0 N/C

Explanation:

Parameters given:

$q_1 = -8.00 * 10^{-7} C$

$x_1 = 6.00 cm$

$q_2 = +8.00 * 10^{-7} C$

$x_2 = 21 cm$

The distance between $q_1$ and $q_2$ is

$21 - 6 = 15cm$

Electric field is given as

$E = \frac{kq}{r^2}$

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to $q_1$ is:

$E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}$

$E_1$ = $-1.28 * 10^6 N/C$

The electric field at the midpoint due to $q_2$ is:

$E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}$

$E_2$ = $1.28 * 10^6 N/C$

The net electric field will be:

E = $E_1$ + $E_2$

E = $-1.28 * 10^6$ + $1.28 * 10^6$

E = 0 N/C

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