A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these charges, in units of N/C?

Question

asked Jul 7, 2020 10.1k views

1 Answer

Uhlen · Answer 1 · 2020-07-11T19:18:47+0000

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC =
71* 10^(- 9) C

Q' = + 42 nC =
42* 10^(- 9) C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = (Q)/(4\pi\epsilon_(o)((d)/(2))^(2))$

$\vec{E} = (71* 10^(- 9))/(4\pi\8.85* 10^(- 12)((1.9)/(2))^(2))$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = (Q')/(4\pi\epsilon_(o)((d)/(2))^(2))$

$\vec{E'} = (42* 10^(- 9))/(4\pi\8.85* 10^(- 12)((1.9)/(2))^(2))$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_(net)} = \vec{E} - \vec{E'}$

$\vec{E_(net)} = 708.03 - 418.84 = 289.19 N/C$