saintv
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Determining Direction of Electric Field! :)
Calculate the electric field at one corner of a square 50 cm on a side if the other three corners are occupied by 250 x 10^-7 C charges.
E = (kQ)/(r^2)
Where:
E = Electric Field Intensity in (N/C)
k = Electrostatics Constant (9 X 10^9 Nm^2/C^2)
Q = Charge in (C)
r = separation in (m)
Firstly, I calculated the E from one corner to each of the three charges:
E1 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.50m)^2
= 9 x 10^5 N/C (Not Sure of Direction)
E2 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.50m)^2
= 9 x 10^5 N/C (Not Sure of Direction)
E3 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.71m)^2
= 4.5 x 10^5 N/C (Not Sure of Direction)
(Note the change in r, as this is diagonal from the corner)
I then added the E.s up:
Etotal = E1 + E2 + E3
= (9 x 10^5 N/C) + (9 x 10^5 N/C) + (4.5 x 10^5 N/C)
Because E1 + E2 are not on the same plane, I would have to use Pythagoras to add the two, and then add it to E3. So,
Etotal = E1 + E2 + E3
= (9 x 10^5 N/C) + (9 x 10^5 N/C) + (4.5 x 10^5 N/C)
= (1.3 x 10^6 N/C) + (4.5 x 10^5 N/C)
= 1.8 X 10^6 N/C
Okay, wait. When I typed it out here and did all the calculations, I got the right answer! That's weird, considering every other time I did it it was wrong.
Anyway, I still have to determine the direction of the Electric Field.
In my answer key, it says: [AWAY FROM CENTER], which I do not understand. Is it because I am dealing with all positive charges, so they all repel each other?
I am simply terrible when it comes to determining direction, so please help!
Homework Statement
Calculate the electric field at one corner of a square 50 cm on a side if the other three corners are occupied by 250 x 10^-7 C charges.
Homework Equations
E = (kQ)/(r^2)
Where:
E = Electric Field Intensity in (N/C)
k = Electrostatics Constant (9 X 10^9 Nm^2/C^2)
Q = Charge in (C)
r = separation in (m)
The Attempt at a Solution
Firstly, I calculated the E from one corner to each of the three charges:
E1 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.50m)^2
= 9 x 10^5 N/C (Not Sure of Direction)
E2 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.50m)^2
= 9 x 10^5 N/C (Not Sure of Direction)
E3 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.71m)^2
= 4.5 x 10^5 N/C (Not Sure of Direction)
(Note the change in r, as this is diagonal from the corner)
I then added the E.s up:
Etotal = E1 + E2 + E3
= (9 x 10^5 N/C) + (9 x 10^5 N/C) + (4.5 x 10^5 N/C)
Because E1 + E2 are not on the same plane, I would have to use Pythagoras to add the two, and then add it to E3. So,
Etotal = E1 + E2 + E3
= (9 x 10^5 N/C) + (9 x 10^5 N/C) + (4.5 x 10^5 N/C)
= (1.3 x 10^6 N/C) + (4.5 x 10^5 N/C)
= 1.8 X 10^6 N/C
Okay, wait. When I typed it out here and did all the calculations, I got the right answer! That's weird, considering every other time I did it it was wrong.
Anyway, I still have to determine the direction of the Electric Field.
In my answer key, it says: [AWAY FROM CENTER], which I do not understand. Is it because I am dealing with all positive charges, so they all repel each other?
I am simply terrible when it comes to determining direction, so please help!