Question

Two charged particles are attached to an x axis: Particle 1 of charge ????2.00 ???? 10????7 C is at position x ???? 6.00 cm and particle 2 of charge ????2.00 ???? 10????7 C is at position x ???? 21.0 cm. Midway between the particles, what is their net electric field in unit-vector notation

Answer 1

Answer:

-639288.89 N/C

Explanation:

x = 21-6 = 13.5 cm

$x_1$ = 6 cm

$x_2$ = 21 cm

$q_1=q_2=-2\times 10^{-7}\ C$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

The net electric field is given by

$E_n=E_1+E_2\\\Rightarrow E_n=-\dfrac{k|q_1|}{(x-x_1)^2}\hat i-\dfrac{k|q_2|}{(x_2-x)^2}\hat i\\\Rightarrow E_n=-\dfrac{8.99\times 10^{9}\times 2\times 10^{-7}}{(0.135-0.06)^2}\hat i-\dfrac{8.99\times 10^{9}\times 2\times 10^{-7}}{(0.21-0.135)^2}\hat i\\\Rightarrow E_n=-639288.89\ N/C(\hat i)$

The net electric field is $-639288.89\ N/C(\hat i)$