What is the electric field at the point?

AI Thread Summary
The discussion focuses on calculating the electric field at point A, located at the upper left corner of a square, due to two equal charges of 7.48 µC placed at the lower corners. The magnitude of the electric field has been correctly calculated as 5.49e+5 N/C using the formula kq/r^2. However, the user is struggling with determining the direction of the electric field. It is suggested that the user consider the contributions from both charges and utilize ratios of side lengths instead of trigonometric functions to find the force components. The conversation emphasizes the importance of accounting for both charges to accurately determine the electric field's direction.
Ammora
Messages
17
Reaction score
0

Homework Statement



Two tiny objects with equal charges of 7.48 µC are placed at the two lower corners of a square with sides of 0.350 m, as shown.

Image: http://www.webassign.net/grrphys2/16-p-039.gif

Find the electric field at point A, the upper left corner. (Assume the positive x-direction points to the right.)
magnitude: ____N/C
direction: ____° CCW from the positive x-axis

Homework Equations



kq/r^2

The Attempt at a Solution



I already got the magnitude right, here's what I did:
kq/r^2
= 8.99e9 (7.48e-6)/(.350)^2 = 5.49e+5

Where I'm stuck is the direction.
I tried tan theta = (.350) and that was wrong, and I tried tan theta= (.35/2)/.35 and that was wrong also.
Any help would be greatly appreciated. This is due tonight. Thanks in advance.
 
Physics news on Phys.org
Ammora said:
I already got the magnitude right, here's what I did:
kq/r^2
= 8.99e9 (7.48e-6)/(.350)^2 = 5.49e+5

That only accounts for the field due to the charge in the lower left corner. What about the other charge?

Also, because you're dealing with a square with sides parallel to the coordinate axes, you should be able to use ratios of side lengths (similar triangles) in lieu of trig functions to work out the force components, if you wish.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top