Calculate the Electric Field at the given point

AI Thread Summary
The discussion focuses on calculating the electric field created by a -12 nC charge at the point (-5.0 cm, -5.0 cm). The user attempts to determine the electric field vector in component form, noting the importance of using unit vectors. There is confusion regarding the choice of unit vector, specifically why the solution manual uses \(\frac{1}{\sqrt{2}}\) for the direction. The correct unit vector should point away from the origin, leading to the conclusion that it is \(-\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})\). Clarification on these vector components is essential for accurately calculating the electric field.
PyMaster
Messages
2
Reaction score
0

Homework Statement


A -12 nC charge is located at the origin.

What is the electric field at the position (x2,y2)=(−5.0cm,−5.0cm)? Write electric field vector in component form.

Homework Equations


upload_2014-9-27_11-44-48.png


The Attempt at a Solution


Here, r^2 = (-0.05 - 0 )^2 meters squared. q=-12*10^-9 C.
For the unit vector, I chose \hat{i} and \hat{j} for the horizontal and vertical components of the vector. In the solutions manual, they choose a unit vector \frac {1} {\sqrt{2}}. I am not sure how they chose this as their unit vector.

I think they start with unit vectors of 5*\hat{i} and 5*\hat{j} then to convert to unit vectors they divide by the magnitude of each of the vectors (which is the length of the hypotenuse; 5*\sqrt(2).

Can someone verify this, please?
 

Attachments

  • upload_2014-9-27_11-40-55.png
    upload_2014-9-27_11-40-55.png
    518 bytes · Views: 530
Physics news on Phys.org
The point is that ##\hat r## should be a unit vector with the same direction as ##\vec r##. The vector ##\vec a = \hat i + \hat j##, this is not a unit vector since ##\vec a^2 = 2##. Additionally, ##\hat r## should be pointing away from the origin, which in (-5cm, -5cm) means that it is given by ##- \frac 1{\sqrt 2}(\hat i + \hat j)##.
 
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