Magnitude of force on charge.

[rcmango]

Homework Statement

A charge of q = +7.50 µC is located in an electric field.

The x and y components of the electric field are Ex = 6.40 103 N/C and Ey = 8.80 103 N/C, respectively.

(a) What is the magnitude of the force on the charge?
in Newtons

(b) Determine the angle that the force makes with the +x axis.
in degrees°

Homework Equations

The Attempt at a Solution

I'm not sure how to use the x and y components to solve this.

not sure why i couldn't just use an equation like this: force per coulomb of charge is F/q0 = N/C

 

[rock.freak667]

Use F=EQ to find the forces in the x and y components. Then it is just basic vector addition

 

[Moetasim]

I can provide a response to this content by explaining the concept of vector addition. In this case, the electric field has both x and y components, which means that it can be represented as a vector in two dimensions. The force on a charge in an electric field is given by the equation F = qE, where q is the charge and E is the electric field. However, since the electric field in this case has both x and y components, we need to use vector addition to find the total electric field.

To find the magnitude of the force on the charge, we can use the Pythagorean theorem. The magnitude of the total electric field is given by E = √(Ex^2 + Ey^2). Plugging in the values given in the problem, we get E = √(6.40x10^3)^2 + (8.80x10^3)^2) = 10.74x10^3 N/C.

Now, we can plug this value into the equation F = qE to find the magnitude of the force. F = (7.50x10^-6 C)(10.74x10^3 N/C) = 80.55x10^-3 N. Therefore, the magnitude of the force on the charge is 80.55x10^-3 N.

To determine the angle that the force makes with the +x axis, we can use the trigonometric relationship for the tangent of an angle. The tangent of an angle is given by tanθ = Opposite/Adjacent. In this case, the opposite side is Ey and the adjacent side is Ex. Therefore, tanθ = Ey/Ex = (8.80x10^3 N/C)/(6.40x10^3 N/C) = 1.375. We can then use a calculator to find the inverse tangent of 1.375, which is approximately 54.46°. Therefore, the angle that the force makes with the +x axis is 54.46°.