Electrostactics- honors physics problem

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Creative-Amy
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Homework Statement



Four point charges of equal magnitude q = 6.4 X 10-19 C are placed at the corners of a square with side length a = 10 cm. Calculate the electric field at P, the center of the square, if q1 and q4 are positive and q2 and q3 are negative. (Use GUESS method.) Answer E = 3.26 X 10-6 N/C


Homework Equations



E=Fe/q E=kQ/r2 Fe=kq1q2/r2 Q=Ne F=ma


k=9.0E9
e=1.6E-19


The Attempt at a Solution



Q=Ne
Q=1*1.6E-19

9E9*1.6E-19/(.10^2)

And now I'm confused...I don't know what steps to follow, can anyone help?
 
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Creative-Amy said:

Homework Statement



Four point charges of equal magnitude q = 6.4 X 10-19 C are placed at the corners of a square with side length a = 10 cm. Calculate the electric field at P, the center of the square, if q1 and q4 are positive and q2 and q3 are negative. (Use GUESS method.) Answer E = 3.26 X 10-6 N/C


Homework Equations



E=Fe/q E=kQ/r2 Fe=kq1q2/r2 Q=Ne F=ma


k=9.0E9
e=1.6E-19


The Attempt at a Solution



Q=Ne
Q=1*1.6E-19

9E9*1.6E-19/(.10^2)

And now I'm confused...I don't know what steps to follow, can anyone help?
Q is not 1.6e-19, that is for an electron or proton. Q is the value given in the problem statement.

Okay, start by calculating the electric field, at the center of the square, due to just one of the charges.
 
equations

Fe=kq1q2/r^2
resultant=sqroot(x^2+y^2)


q4-q1
y component

((9E9)(6.4E-19)^2)/.1^2 = 3.6864E-25

q3-q4
x component

((9E9)(6.4E-19)^2)/.1^2 = 3.6864E-25

Solve for resultant

sqroot( 3.6864E-25^2+3.6864E-25^2) = 5.21E-25 = Fe

equation

E=Fe/q

q=1.6E-19

5.21E-25/1.6E-19 = 3.258E-6

Answer= 3.26E-6

Correct?