Four point charges are placed at the corners of a square. Each side of the square has length 2 m. Determine the magnitude of the electric field at the point P (center of the square)?
A) 2 * 10^-6 N/C
B) 3 * 10^-6 N/C
C) 9 * 10 ^3 N/C
D) 1.8 * 10^4 N/C
E) 2.7*10^4 N/C
What is the magnitude of the electric field due to a 4 * 10 ^-6 C charge at a point 0.08 m away?
A) 1800 N/C
B) 1 *10^5 N/C
C) 5.63 *10^6 N/C
D) 9 * 10^4 N/C
E) 3.6 * 10^6 N/C
*Help, I need to know this for the exam review!
2006-12-23 13:00:46 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The electric field strength is the vector sum of the individual fields. The web site below is really good for computing the field as a result of 1 or 2 charges, and shows that "C)" is the answer to the second question.
It also helps add up to two points, which I think you should be able to extend to 4 by hand.
2006-12-23 13:28:05 · answer #1 · answered by firefly 6 · 0⤊ 0⤋
using the formula E = (a million/4piE°)(q/d^2) the place E° is absoute permitivity of loose area d= 0.25m q = 6.0 × 10–9 C (a million/4piE°)=(9*10^9) as a result electric field= (9*10^9)*(6*10^-9)/(0.25)^2 = 864 N/C
2016-11-23 14:05:24 · answer #2 · answered by ? 4 · 0⤊ 0⤋
I don't have the slightest idea..just being honest
2006-12-23 13:52:50 · answer #3 · answered by Leigh P 3 · 0⤊ 0⤋