(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two tiny objects with equal charges of 65.0 µC are placed at two corners of a square with sides of 0.335 m, as shown. How far above and to the left of the corner of the square labeled A would you place a third small object with the same charge so that the electric field is zero at A?

above corner A= ?m

to the left of corner A= ?m

image http://www.webassign.net/grr/p16-27.gif

2. Relevant equations

Coulomb's Law: F = (k * q1 * q2 ) / r^2

Electric Field: = k*q / r^2

3. The attempt at a solution

.287933617

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Force from just below A.

d = 0.335 m

k = 9*10^9

q1 = 65 *10^-6 C

E = kq1/d^2

E = 9*10^9*65*10^-6/(0.335)^2

E =5212752.26 N/q

Force from charge on the same diagronal as A

==================================

d = sqrt(0.335^2 + 0.335^2) A diagronal of a square = sqrt(side^2 + side^2)

d = 0.4737 m

E2 = k*q1/d

E2 = 9*10^9 * 65 * 10^-6/(0.4737)^2

E2 = 2607048.416 N/q

Summing the Fields.

Vertical

Total Vertical = 5212752.26 + 2607048.416 *Sin(45o)

the angle a diagonal makes with with vertical is 45o

Total Vertical = 5212752.26 N/q + 1843461.614 N/q

Total Vertical = 7056213.874 N/q

Total Horizontal = 2607048.416 * cos(45o)

Total Horizontal = 1843461.614 N/q

I used the following equation for the rest, but im not getting the right answer, please if anyone can help me, this is due in an hour. Thank you so much ive been trying this for hours

d^2 = kq1/E

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# Finding charge so that the electric field is zero

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