A 1500 kg truck travelling north at 60 km/hr collides with a 1200 kg car moving east at 15km/hr. If the two cars remain locked t
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To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.
The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:
F = Eq
F is the electric force, E is the electric field, and q is the charge
Let us adjust the equation for only the horizontal components of the above quantities:
Fx = (Ex)(q)
Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.
Given values:
F = 6.57×10⁻²N
q = 6.80×10³C
Plug in these values and solve for Ex:
6.57x10⁻² = Ex(6.80×10³)
<u>Ex = 9.66×10⁻⁶N/C</u>
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Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.
We know the weight of the sphere is given by:
W = mg
W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.
We also know this equation:
F = Eq
Let us adjust for the vertical components:
Fy = (Ey)(q)
Set Fy equal to W and we get:
(Ey)(q) = mg
Given values:
q = 6.80×10³C
m = 0.018kg
g = 9.81m/s²
Plug in the values and solve for Ey:
(Ey)(6.80×10³) = 0.018(9.81)
<u>Ey = 2.60×10⁻⁵N/C</u>
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Let's now use the Pythagorean theorem to find the total magnitude of the electric field:
E =
E = 2.77×10⁻⁵N/C
The direction of the electric field is given by:
θ = tan⁻¹(Ey/Ex)
θ = 20.4° off the horizontal