Originally Posted by **Ess J** q1 .______________________.q2
0.1meters
The above fig. shows 2 point charges q1 (1x10-6 C) and q2 (2x10-6 C) are 0.1m away. On which point on the joining line is the Electrin Field Steangth zero??
(the smaller '-6's r meant to be the powers of 10) |

The electric field a distance r from a point particle of charge q is given by

$\displaystyle E = \frac{k q}{r^2}$

or

$\displaystyle E = \frac{1}{4 \pi \epsilon _0} \frac{q}{r^2}$

depending on how your book does it. The field is a vector field and points away from a positive charge and toward a negative charge.

Find the field due to each point charge and add them vectorally. Putting the origin at q1 and the positive x direction toward q2 we then have at a distance x from q1:

$\displaystyle E = \frac{k q_1}{x^2} - \frac{k q_2}{(0.1 - x)^2}$

Set E = 0 and solve for x.

-Dan