A positive charge of 1.0*10^-5 C, shown in the picture link, experiences a force of 0.30 N when it is located at a certain point. What is the electric field intensity at that point?
*picture link: http://img340.imageshack.us/img340/4786/physics67bu5.jpg
Please help me with this problem.
I don't understand it.
2007-04-24 12:36:21 · 3 answers · asked by vicky p 1 in Science & Mathematics ➔ Physics
The force on a charge in an electric field is q*E, where q is the charge in coulombs, and E in volts/meter, force in Newtons. Therefore, E = F/q = 0.3/1.0*10^-5 =3*10^4 volt/meter
2007-04-24 12:48:46 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
electric field is in Newtons per Coulomb, so just divide N by C to find E
2007-04-24 19:49:43 · answer #2 · answered by hello 6 · 0⤊ 0⤋
E = F / q
E = .30 / (1 x 10^-5)
E = 30,000 N/C
2007-04-24 19:51:27 · answer #3 · answered by 7 · 0⤊ 0⤋