Posted by **Joanna Luz** on Monday, January 24, 2011 at 12:36am.

pls help me solve this problem..

charges of +2.0, 3.0 and -8.0µC are placed at the vertices of an equilateral triangle of a side 10 cm. Calculate the magnitude of the force acting on the -8µC charge due to the other two charges.

- physics -
**drwls**, Monday, January 24, 2011 at 6:22am
Add the vector sum of

-k*2*8*10^-12/0.1^2 and

-k*3*8*10^-12/0.1^2,

applied 60 degrees apart.

k is the Coulomb constant,

8.99*10^9 N/m^2*C^2

There will be an attraction force in a direction between the two other points of the triangle. It will not bisect the angle, because the forces are unequal.

## Answer this Question

## Related Questions

- physics - the side of an equilateral triangle is 20 cm identical charges of 1.5x...
- Physics - Three identical charges of 2.0 uC are placed at the vertices of an ...
- physics - three point charges each of magnitude +q are located at the vertices ...
- physics - Three charges of 5 x 10^(-6) C are placed on the vertices of an ...
- physics - Three equal charges are placed at the corners of an equilateral ...
- physics - Three charged particles are placed at the corners of an equilateral ...
- Physics - three identical 2*`0^-5C point charges are placed a the corners of an ...
- Physics - Three point charges + q, -q and + q are placed at the vertices P, Q ...
- Electr. and Magnetism - 1) A charge of 80nC is uniformly distributed along the x...
- physics - Three charged particles are placed at the corners of an equilateral ...

More Related Questions