Solving for Electric Field at an Angle: Ex = 10.0V/m

In summary, the conversation discusses the calculation of the electric field (Ex) at a point (d) that is 5.00m away from the origin, where the x-coordinate is 4.00m and the y-coordinate is 3.00m. The formula used is Ex = -dV/dx, with the given value for Ex being 10.0Vm-1. By substituting the values, the solution comes out to be -50.0V for dV. However, the given solution is at 50.5V, which may be due to the field being directed at an angle of 45.0 degrees from the origin, which was not taken into account in the calculation. This emphasizes
  • #1
TTauri
4
1
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= d2=x2+y2
x=4.00m
y=3.00m
d2=4.002+3.002
d=5.00m
Ex = -dV/dx
10.0Vm-1=-dV/5.00m
10.0Vm-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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  • #2
Realize that direction matters. If you drew a line from the origin to the point in question, what angle would the field make with that line?
 

Related to Solving for Electric Field at an Angle: Ex = 10.0V/m

1. What is the equation for calculating electric field at an angle?

The equation for calculating electric field at an angle is Ex = (kq/r^2)cosθ, where Ex is the electric field strength, k is the Coulomb's constant, q is the charge of the object creating the field, r is the distance between the object and the point where the field is being measured, and θ is the angle between the object and the point where the field is being measured.

2. How do you determine the direction of the electric field at an angle?

The direction of the electric field at an angle is determined by the direction of the force that would be exerted on a positive test charge placed at that point. If the angle is 0 degrees, the electric field will point directly away from the charged object. If the angle is 180 degrees, the electric field will point directly towards the charged object.

3. What does a positive or negative electric field at an angle indicate?

A positive electric field at an angle indicates that the electric field is pointing away from the charged object. This means that a positive test charge placed at that point would experience a repulsive force. A negative electric field at an angle indicates that the electric field is pointing towards the charged object, and a positive test charge placed at that point would experience an attractive force.

4. How does the distance between the object and the point of measurement affect the electric field at an angle?

According to the equation for electric field at an angle, the strength of the electric field is inversely proportional to the square of the distance between the object and the point of measurement. This means that as the distance increases, the electric field strength decreases, and vice versa.

5. Can the electric field at an angle ever be equal to zero?

Yes, the electric field at an angle can be equal to zero. This would occur if the angle between the object and the point of measurement is 90 degrees, since the cosine of 90 degrees is equal to 0. In this case, there would be no component of the electric field in the horizontal direction, resulting in a net electric field of 0.

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