- #1

TTauri

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- Homework Statement
- In a certain region of space, there is a uniform electric field of magnitude

10.0 V m−1 directed at 45.0◦ to the positive direction of both the x and y axes. (There is no z

component of the electric field.) The electric potential at the origin is +100 V. What is the electric

potential at the point where x = 4.00 m and y = 3.00 m?

- Relevant Equations
- Ex = -dV/dx

Ex =10.0Vm-1

dx= d

x=4.00m

y=3.00m

d

d=5.00m

Ex = -dV/dx

10.0Vm

10.0Vm

-50.0V=dV

So from origin at 100V-50.0V = 50.0V

But the solution I am given gives the answer at 50.5V and the information "directed at 45.0◦ " does not seem to have been used so may related to the difference.

I have hunted a bit round the internet and through textbooks but cannot find this formulation with the field being directed at an angle.

dx= d

^{2}=x^{2}+y^{2}x=4.00m

y=3.00m

d

^{2}=4.00^{2}+3.00^{2}d=5.00m

Ex = -dV/dx

10.0Vm

^{-1}=-dV/5.00m10.0Vm

^{-1}*5.00m=-dV-50.0V=dV

So from origin at 100V-50.0V = 50.0V

But the solution I am given gives the answer at 50.5V and the information "directed at 45.0◦ " does not seem to have been used so may related to the difference.

I have hunted a bit round the internet and through textbooks but cannot find this formulation with the field being directed at an angle.

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