Find Net Electric Field Magnitude & Direction in an Equilateral Triangle

AI Thread Summary
To find the net electric field at the centroid of an equilateral triangle, use the equation for electric field, E = kq/r^2, where k is Coulomb's constant, q is the charge, and r is the distance from the charge to the point of interest. Since the question involves direction, it's essential to treat the electric field as a vector, incorporating both magnitude and direction. The direction can be determined by breaking the electric field into x and y components based on the geometry of the triangle. This approach ensures accurate calculation of the net electric field at the centroid.
kevykevy
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Find the magnitude and direction of the net electric field at the centre (or centroid) of the triangle.

The triangle is equilateral and the point of reference is in the middle of the triangle.

What I want to know is what equation do I use to figure this out?

At first I thought it was: Fnet = kq1q2/r^2

Is that right?, if not can you tell what to use?
 
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That's almost right...the question asks for the net electric field, which means you have to use a vector, not just the magnitude as you've shown.

\vec{E} = \frac{kq}{r^2}\hat{r}
 
Also, how do I find the direction of the field?

Do I have to use some sort of x and y component breakdown?
 
kevykevy said:
Do I have to use some sort of x and y component breakdown?

That's right.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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